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  • ACM模板

    矩阵快速幂

    参考题目:https://codeforces.com/problemset/problem/678/D

    #include<bits/stdc++.h>
    #define ios std::ios::sync_with_stdio(false)
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,n,a) for (int i=n;i>=a;i--)
    #define int long long
    #define ll long long
    #define MOD 1000000007
    using namespace std;
    const int MAXN = 10;
    struct Matrix
    {
        ll mat[MAXN][MAXN];
    };
    Matrix P; 
    Matrix I; //单位矩阵
    Matrix Mul_Matrix(Matrix a, Matrix b , ll mod)
    {
        Matrix c;
        for(int i = 0 ; i < MAXN ; i ++)
            for(int j = 0 ; j < MAXN ; j ++)
            {
                c.mat[i][j] = 0;
                for(int k = 0 ; k < MAXN ; k ++)
                {
                    c.mat[i][j] += (a.mat[i][k] * b.mat[k][j]) % mod;
                    c.mat[i][j] %= mod;
                }
            }
        return c;
    }
    Matrix pow_mod_Matrix(Matrix P , ll n , ll mod)
    {
        Matrix ans = I, b = P;
        while(n)
        {
            if(n & 1) ans = Mul_Matrix(ans , b , mod);
            n >>= 1;
            b = Mul_Matrix(b , b , mod);
        }
        return ans;
    }
    void Matrix_init()
    {
        for(int i = 0 ; i < MAXN ; i ++) I.mat[i][i] = 1;
    }
    signed main()
    {
        ios;
        int a , b , n , x;
        cin >> a >> b >> n >> x;
        Matrix_init();
        {
            P.mat[0][0] = a, P.mat[0][1] = 0;
            P.mat[1][0] = 1, P.mat[1][1] = 1;
            Matrix res = pow_mod_Matrix(P , n , MOD);
            ll ans = x * res.mat[0][0] + b * res.mat[1][0];
            ans = (ans % MOD + MOD) % MOD;
            cout << ans << '
    ';
        }
        return 0;
    }
    

    Dijkstra堆优化

    参考题目:https://www.luogu.com.cn/problem/P4779

    #include<bits/stdc++.h>
    #define int long long
    #define rep(i , a , b) for(int i = a ; i <= b ; i ++)
    using namespace std;
    const int inf = 0x3f3f3f3f;
    const int N = 2e5 + 10;
    struct node{
        int dis , pos;
        bool operator <( const node &x )const{
            return x.dis < dis;
        }
    };
    struct Edge{
        int to , dis , nex;
    }edge[N << 1];
    int head[N], dis[N] , tot , vis[N] , n , m , s;
    inline void add_edge(int u , int v , int d)
    {
        edge[++ tot].dis = d;
        edge[tot].to = v;
        edge[tot].nex = head[u];
        head[u] = tot;
    }
    inline void dijkstra(int s)
    {
        memset(dis , inf , sizeof(dis)) ;
        dis[s] = 0;
        priority_queue <node> que;
        que.push(node{0, s});
        while(!que.empty())
        {
            node tmp = que.top();
            que.pop();
            int u = tmp.pos, d = tmp.dis;
            if(vis[u]) continue;
            vis[u] = 1;
            for(int i = head[u] ; i ; i = edge[i].nex)
            {
                int v = edge[i].to;
                if(dis[v] > dis[u] + edge[i].dis)
                {
                    dis[v] = dis[u] + edge[i].dis;
                    if(!vis[v]) que.push(node{dis[v] , v});
                }
            }
        }
    }
    signed main()
    {
        ios::sync_with_stdio(false);
        cin >> n >> m >> s;
        rep(i , 1 , m)
        {
            int u , v , w;
            cin >> u >> v >> w;
            add_edge(u , v , w);
        }
        dijkstra(s);
        rep(i , 1 , n) cout << dis[i] << " ";
        cout << '
    ';
        return 0;
    }
    

    Dijkstra配对堆

    #include<bits/stdc++.h>
    #include<ext/pb_ds/priority_queue.hpp>
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,n,a) for (int i=n;i>=a;i--)
    #define il inline
    using namespace std;
    using namespace __gnu_pbds;
    const int inf (0x3f3f3f3f);
    typedef __gnu_pbds::priority_queue< pair<int,int> ,greater< pair<int,int> >,pairing_heap_tag > heap;
    template<typename T>void read(T &res){bool flag=false;char ch;while(!isdigit(ch=getchar()))(ch=='-')&&(flag=true);
    for(res=ch-48;isdigit(ch=getchar());res=(res<<1)+(res<<3)+ch - 48);flag&&(res=-res);}
    template<typename T>void Out(T x){if(x<0)putchar('-'),x=-x;if(x>9)Out(x/10);putchar(x%10+'0');}
    const int N = 2e5 + 10;
    heap que;
    heap::point_iterator id[N];
    struct Edge{
        int nex , to , w;
    }edge[N << 1];
    int head[N] , tot;
    il void add_edge(int u , int v , int w)
    { 
        edge[++ tot].nex = head[u];
        edge[tot].to = v;
        edge[tot].w = w;
        head[u] = tot;
    }
    int n , m , s , dis[N];
    il void dij(int st)
    {
        memset(dis , inf , sizeof(dis));
        dis[st] = 0;
        id[st] = que.push(make_pair(0 , st));
        while(!que.empty())
        {
            int u = que.top().second;
            que.pop();
            for(int i = head[u] ; i ; i = edge[i].nex)
            {
                int v = edge[i].to ;
                int w = edge[i].w ;
                if(w + dis[u] < dis[v])
                {
                    dis[v] = dis[u] + edge[i].w;
                    if(id[edge[i].to] != 0) que.modify(id[v] , make_pair(dis[v] , v));
                    else id[v] = que.push(make_pair(dis[v] , v));
                }
            }
        }
    }
    signed main()
    {
        read(n) , read(m) , read(s); 
        rep(i , 1 , m)
        {
            int u , v , w;
            read(u) , read(v) , read(w);
            add_edge(u , v , w);
        }
        dij(s);
        rep(i , 1 , n) Out(dis[i]) , putchar(' ');
        return 0;
    }
    

    Manacher

    参考题目:https://www.luogu.com.cn/problem/P3805

    #include<bits/stdc++.h>
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,n,a) for (int i=n;i>=a;i--)
    using namespace std;
    const int N = 3e7 + 10;
    int pa[N]; 
    string a ;
    int Manacher(string a , int *p)
    {
        string t = "$#";
        for(auto i : a) t += i , t += '#';
        int mx = 0 , id = 0 ;
        int len = t.size() , ans = 0; 
        for(int i = 1 ; i < len ; i ++)
        {
            p[i] = mx > i ? min(p[2 * id - i] , mx - i) : 1;
            while(t[i + p[i]] == t[i - p[i]]) p[i] ++ ;
            if(mx < i + p[i]) mx = i + p[i] , id = i;
            ans = max(ans , p[i] - 1);
        }
        return ans;
    }
    signed main()
    {
        ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0);
        cin >> a ;
        cout << Manacher(a , pa) << '
    ';
        return 0;
    } 
    

    主席树

    参考题目:https://www.luogu.com.cn/problem/P3834

    #include<bits/stdc++.h>
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    using namespace std;
    const int N = 3e5 + 10;
    struct Chairman_Tree{
        int l , r , sum;
    }Ctree[N << 5]; 
    vector<int>vec;
    int root[N] , a[N] , Ctot;
    void update(int l , int r , int pre , int &now , int pos)
    {
        Ctree[++ Ctot] = Ctree[pre]; 
        Ctree[Ctot].sum ++ ;
        now = Ctot;
        if(l == r) return ;
        int mid = l + r >> 1;
        if(pos <= mid) update(l , mid , Ctree[pre].l , Ctree[now].l , pos);
        else update(mid + 1 , r , Ctree[pre].r , Ctree[now].r , pos);
    }
    int query(int l , int r , int L , int R , int k)
    {
        if(l == r) return l ;
        int mid = l + r >> 1;
        int cha = Ctree[Ctree[R].l].sum - Ctree[Ctree[L].l].sum;
        if(cha >= k) return query(l , mid , Ctree[L].l , Ctree[R].l , k);
        else return query(mid + 1 , r , Ctree[L].r , Ctree[R].r , k - cha);
    }
    int get_id(int x)
    {
        return lower_bound(vec.begin() , vec.end() , x) - vec.begin() + 1;
    }
    signed main()
    {
        ios::sync_with_stdio(false);
        int n , m ;
        cin >> n >> m;
        rep(i , 1 , n)
        {
            cin >> a[i];
            vec.push_back(a[i]);
        }
        sort(vec.begin() , vec.end());
        vec.erase(unique(vec.begin() , vec.end()) , vec.end()) ;
        rep(i , 1 , n)
            update(1 , n , root[i - 1] , root[i] , get_id(a[i]));
        while(m --)
        {
            int l , r , k;
            cin >> l >> r >> k ; 
            int pos = query(1 , n , root[l - 1] , root[r] , k);
            cout << vec[pos - 1] << '
    ' ; 
        }
        return 0;
    }
    

    二次剩余

    *参考题目:https://www.luogu.com.cn/problem/P5491*

    #include<bits/stdc++.h>
    using namespace std;
    #define ll long long
    ll w;
    struct num
    {
        ll x, y;
    };
    num mul(num a, num b, ll p)
    {
        num ans = {0, 0};
        ans.x = ((a.x * b.x % p + a.y * b.y % p * w % p) % p + p) % p;
        ans.y = ((a.x * b.y % p + a.y * b.x % p) % p + p) % p;
        return ans;
    }
    ll powwR(ll a, ll b, ll p)
    {
        ll ans = 1;
        while (b)
        {
            if (b & 1)
                ans = 1ll * ans % p * a % p;
            a = a % p * a % p;
            b >>= 1;
        }
        return ans % p;
    }
    ll powwi(num a, ll b, ll p)
    {
        num ans = {1, 0};
        while (b)
        {
            if (b & 1)
                ans = mul(ans, a, p);
            a = mul(a, a, p);
            b >>= 1;
        }
        return ans.x % p;
    }
    ll M_sqrt(ll n, ll p)
    {
        n %= p;
        if (p == 2)
            return n;
        if (powwR(n, (p - 1) / 2, p) == p - 1)
            return -1; //不存在
        ll a;
        while (1)
        {
            a = rand() % p;
            w = ((a * a % p - n) % p + p) % p;
            if (powwR(w, (p - 1) / 2, p) == p - 1)
                break;
        }
        num x = {a, 1};
        return powwi(x, (p + 1) / 2, p);
    }
    signed main()
    {
        ios::sync_with_stdio(false);
        int t;
        cin >> t;
        while(t --)
        {
            ll n , p;
            cin >> n >> p ;
            if(!n)
            {
                cout << 0 << '
    ' ; 
                continue ;
            }
            ll ans1 = M_sqrt(n , p) , ans2;
            if(ans1 == -1) cout << "Hola!
    ";
            else
            {
                ans2 = p - ans1;
                if(ans1 > ans2)    swap(ans1 , ans2);
                if(ans1 == ans2) cout << ans1 << '
    '; 
                else cout << ans1 << " " << ans2 << '
    ' ;
            }
        }
        return 0;
    }
    

    KMP

    参考题目:https://www.luogu.com.cn/problem/P3375

    #include<bits/stdc++.h>
    using namespace std;
    const int N = 1e6 + 10;
    int nex[N] ; 
    vector<int>ans;
    string s ,  t;
    void get_nex(string s , int *nex)
    {
        int i = 0 , j = -1 , len = s.size();
        nex[i] = j;
        while(i < len)
        {
            while(j != -1 && s[i] != s[j]) j = nex[j];
            nex[++ i] = ++ j ;
        }
    }
    void KMP(string s , string t)
    {
        get_nex(t , nex) ; 
        int i = 0 , j = 0 , lens = s.size() , lent = t.size();    
        while(i < lens)
        {
            while(j != -1 && s[i] != t[j]) j = nex[j];
            i ++ , j ++ ;
            if(j == lent) ans.push_back(i - lent + 1);
        }
    }
    signed main()
    {
        ios::sync_with_stdio(false);
        cin >> s >> t ;
        KMP(s , t);
        for(auto i : ans) cout << i << '
    ' ; 
        for(int i = 1 ; i <= t.size() ; i ++ ) cout << nex[i] << " ";
        cout << '
    ';
         return 0;
    }
    

    MatrixTree矩阵树——求无向图生成树个数

    参考题目:https://www.luogu.com.cn/problem/SP104

    #include<bits/stdc++.h>
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,n,a) for (int i=n;i>=a;i--)
    #define int long long
    #define ll long long
    using namespace std;
    const int N = 2e2 + 10;
    const double eps = 1e-12;
    int G[N][N];
    int n , m;
    int dcmp(int x)
    {
       if(x <= eps || x >= -eps) return 0;
       else return x < 0 ? -1 : 1;
    }
    void init(int n)
    {
       rep(i , 0 , n) rep(j , 0 , n) G[i][j] = 0;
    }
    ll Gauss(int n)
    {
       ll ans = 1;
       rep(i , 1 , n - 1)
       {
           rep(k , i + 1 , n - 1)
           {
               while(G[k][i])
               {
                   int d = G[i][i] / G[k][i];
                   rep(j , i , n - 1) G[i][j] = (G[i][j] - 1LL * d * G[k][j]);
                   swap(G[i] , G[k]) , ans = -ans;
               }
           }
           ans = 1LL * ans * G[i][i];
       }
       return ans;
    }
    signed main()
    {
       ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0);
       int T = 1;
       cin >> T;
       while(T --)
       {
           cin >> n >> m;
           init(n);
           rep(i , 1 , m)
           {
               int x , y ;
               cin >> x >> y ;
               G[x][x] ++ , G[y][y] ++;
               G[x][y] -- , G[y][x] --;
           }
           int tot = Gauss(n);
           cout << tot << '
    ';
       }
       return 0;
    }
    

    AC自动机—fail树

    参考题目:https://www.luogu.com.cn/problem/P5357

    #include<bits/stdc++.h>
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,n,a) for (int i=n;i>=a;i--)
    #define int long long
    #define ll long long
    using namespace std;
    const int N = 3e5 + 10;
    struct Edge{
        int nex , to;
    }edge[N << 1];
    int head[N] , TOT;
    void add_edge(int u , int v)
    {
        edge[++ TOT].nex = head[u] ;
        edge[TOT].to = v;
        head[u] = TOT;
    }
    int cnt[N] , fail[N] , num[N];
    int tree[N][26];
    int tot = 0;
    void del(int x)
    {
        cnt[x] = 0 , fail[x] = 0;
        rep(i , 0 , 25) tree[x][i] = 0;
    }
    void insert(string s , int id)
    {
        int root = 0;
        for(auto i : s)
        {
            int x = i - 'a';
            if(!tree[root][x]) 
            {
                tree[root][x] = ++ tot;
                del(tot);
            }
            root = tree[root][x]; 
        }
        num[id] = root; 
    }
    void build()
    {
        queue<int>que;
        rep(i , 0 , 25) if(tree[0][i])
        {
            int v = tree[0][i];
            fail[v] = 0;
            que.push(v);
        }
        while(!que.empty())
        {
            int u = que.front(); 
            que.pop();
            rep(i , 0 , 25)
            {
                if(tree[u][i])
                {
                    int v = tree[u][i];
                    fail[v] = tree[fail[u]][i];
                    que.push(v);
                }
                else tree[u][i] = tree[fail[u]][i];
            }
        }
    }
    void query(string s)
    {
        int p = 0 , res = 0;
        for(auto i : s)
        {
            int x = i - 'a';
            p = tree[p][x] ;
            cnt[p] ++ ;
        }
    }
    string st[N] , s;
    void dfs(int u , int far)
    {
        for(int i = head[u] ; i ; i = edge[i].nex)
        {
            int v = edge[i].to ; 
            if(v == far) continue ;
            dfs(v , u);
            cnt[u] += cnt[v]; 
        }
    }
    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0) , cout.tie(0);
        int n , T = 1;
          cin >> n;
          rep(i , 1 , n) cin >> st[i] , insert(st[i] , i);
          build();
          cin >> s;
          query(s);
          rep(i , 1 , tot) add_edge(fail[i] , i);
          dfs(0 , -1);
          rep(i , 1 , n) cout << cnt[num[i]] << '
    ';    
        return 0;
    }
    

    参考题目:https://www.luogu.com.cn/problem/P3796

    #include<bits/stdc++.h>
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,n,a) for (int i=n;i>=a;i--)
    using namespace std;
    const int N = 1e6 + 10;
    int cnt[N] , fail[N] , num[N];
    int tree[N][26];
    int tot = 0;
    struct Edge{
        int nex , to;
    }edge[N << 1];
    int head[N] , TOT;
    void add_edge(int u , int v)
    {
        edge[++ TOT].nex = head[u] ; 
        edge[TOT].to = v;
        head[u] = TOT;
    }
    void del(int x)
    {
        cnt[x] = 0 , fail[x] = 0;
        rep(i , 0 , 25) tree[x][i] = 0;
    }
    void insert(string s , int id)
    {
        int root = 0;
        for(auto i : s)
        {
            int x = i - 'a';
            if(!tree[root][x]) 
            {
                tree[root][x] = ++ tot;
                del(tot);
            }
            root = tree[root][x]; 
        }
        num[id] = root;
    }
    void build()
    {
        queue<int>que;
        rep(i , 0 , 25) if(tree[0][i])
        {
            int v = tree[0][i];
            fail[v] = 0;
            que.push(v);
        }
        while(!que.empty())
        {
            int u = que.front(); 
            que.pop();
            rep(i , 0 , 25)
            {
                if(tree[u][i])
                {
                    int v = tree[u][i];
                    fail[v] = tree[fail[u]][i];
                    que.push(v);
                }
                else tree[u][i] = tree[fail[u]][i];
            }
        }
    }
    int query(string s)
    {
        int p = 0 , res = 0;
        for(auto i : s)
        {
            int x = i - 'a';
            p = tree[p][x] ;
            int fi = p;
            cnt[fi] ++ ;
        }
        return res;
    }
    void dfs(int u , int far)
    {
        for(int i = head[u] ; i ; i = edge[i].nex)
        {
            int v = edge[i].to;
            if(v == far) continue ;
            dfs(v , u);
            cnt[u] += cnt[v];
        }
    }
    string st[N] , s;
    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0) , cout.tie(0);
        int n , T = 1;
        while(cin >> n && n)
        {
            rep(i , 1 , n) cin >> st[i] , insert(st[i] , i);
            build();
            cin >> s;
            query(s);
            rep(i , 1 , tot) add_edge(fail[i] , i);
            dfs(0 , -1);
            int ans = 0;
            rep(i , 1 , n) ans = max(ans , cnt[num[i]]);
            cout << ans << '
    ';
            rep(i , 1 , n) if(ans == cnt[num[i]]) {cout << st[i] << '
    ';}
            del(0);
            tot = 0;
            memset(head , 0 , sizeof(head)) , TOT = 0;
        }
        return 0;
    }
    

    AC自动机—fail指针乱跳版

    参考题目:https://www.luogu.com.cn/problem/P3808

    #include<bits/stdc++.h>
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,n,a) for (int i=n;i>=a;i--)
    using namespace std;
    const int N = 1e6 + 10;
    int cnt[N] , fail[N];
    int tree[N][26];
    int tot = 0;
    string s; 
    void del(int x)
    {
       cnt[x] = 0 , fail[x] = 0;
       rep(i , 0 , 25) tree[x][i] = 0;
    }
    void insert(string s)
    {
       int root = 0;
       for(auto i : s)
       {
           int x = i - 'a';
           if(!tree[root][x]) 
           {
               tree[root][x] = ++ tot;
               del(tot);
           }
           root = tree[root][x]; 
       }
       cnt[root] ++ ;
    }
    void build()
    {
       queue<int>que;
       rep(i , 0 , 25) if(tree[0][i])
       {
           int v = tree[0][i];
           fail[v] = 0;
           que.push(v);
       }
       while(!que.empty())
       {
           int u = que.front(); 
           que.pop();
           rep(i , 0 , 25)
           {
               if(tree[u][i])
               {
                   int v = tree[u][i];
                   fail[v] = tree[fail[u]][i];
                   que.push(v);
               }
               else tree[u][i] = tree[fail[u]][i];
           }
       }
    }
    int query(string s)
    {
       int p = 0 , res = 0;
       for(auto i : s)
       {
           int x = i - 'a';
           p = tree[p][x] ;
           int fi = p;
           while(fi && cnt[fi] != -1)
           {
               res += cnt[fi];
               cnt[fi] = -1;
               fi = fail[fi];
           }
       }
       return res;
    }
    signed main()
    {
       ios::sync_with_stdio(false);
       cin.tie(0) , cout.tie(0);
       int n , T = 1;
       while(T --)
       {
           cin >> n;
           rep(i , 1 , n) cin >> s , insert(s);
           build();
           cin >> s;
           cout << query(s) << '
    ';
           del(0);
           tot = 0;
       }
       return 0;
    }
    

    网络流—预流推进HLPP

    参考题目:https://www.luogu.com.cn/problem/P4722

    #pragma GCC optimize(3,"Ofast","inline")
    #include<bits/stdc++.h>
    #define ll long long
    using namespace std;
    const int inf = INT_MAX;
    const ll INF = (0x3f3f3f3f3f3f3f3fll);
    const int maxn = 1.2e3 + 5;
    struct HLPP
    {
        struct Edge
        {
            int v, rev;
            ll cap;
        };
        int n, sp, tp, lim, ht, lcnt;
        ll exf[maxn];
        vector<Edge> G[maxn];
        vector<int> hq[maxn], gap[maxn], h, sum;
        void init(int nn, int s, int t)
        {
            sp = s, tp = t, n = nn, lim = n + 1, ht = lcnt = 0;
            for(int i = 1; i <= n; ++i) G[i].clear(), exf[i] = 0;
        }
        void add_edge(int u, int v, ll cap)
        {
            G[u].push_back( {v, (int)(G[v].size()) , cap});
            G[v].push_back( {u, (int)(G[u].size()) - 1, 0});
        }
        void update(int u, int nh)
        {
            ++lcnt;
            if(h[u] != lim) --sum[h[u]];
            h[u] = nh;
            if(nh == lim) return;
            ++sum[ht = nh];
            gap[nh].push_back(u);
            if(exf[u] > 0) hq[nh].push_back(u);
        }
        void relabel()
        {
            queue<int> q;
            for(int i = 0; i <= lim; ++i) hq[i].clear(), gap[i].clear();
            h.assign(lim, lim), sum.assign(lim, 0), q.push(tp);
            lcnt = ht = h[tp] = 0;
            while(!q.empty())
            {
                int u = q.front();
                q.pop();
                for(Edge &e : G[u]) if(h[e.v] == lim && G[e.v][e.rev].cap) update(e.v, h[u] + 1), q.push(e.v);
                ht = h[u];
            }
        }
        void push(int u, Edge &e)
        {
            if(!exf[e.v]) hq[h[e.v]].push_back(e.v);
            ll df = min(exf[u], e.cap);
            e.cap -= df, G[e.v][e.rev].cap += df;
            exf[u] -= df, exf[e.v] += df;
        }
        void discharge(int u)
        {
            int nh = lim;
            if(h[u] == lim) return;
            for(Edge &e : G[u])
            {
                if(!e.cap) continue;
                if(h[u] == h[e.v] + 1)
                {
                    push(u, e);
                    if(exf[u] <= 0) return;
                }
                else if(nh > h[e.v] + 1) nh = h[e.v] + 1;
            }
            if(sum[h[u]] > 1) update(u, nh);
            else
            {
                for( ; ht >= h[u]; gap[ht--].clear())
                for(int &i : gap[ht]) update(i, lim);
            }
        }
        ll hlpp()
        {
            exf[sp] = INF, exf[tp] = -INF, relabel();
            for(Edge &e : G[sp]) push(sp, e);
            for( ; ~ht; --ht)
            {
                while(!hq[ht].empty())
                {
                    int u = hq[ht].back();
                    hq[ht].pop_back();
                    discharge(u);
                    if(lcnt > (n << 2)) relabel();
                }
            }
            return exf[tp] + INF;
        }
    }hp;
    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0) , cout.tie(0);
        int n, m, s, t, u, v, w;
        cin >> n >> m >> s >> t;
        hp.init(n, s, t);
        while(m --)
        {
            cin >> u >> v >> w;
            hp.add_edge(u, v, w);
        }
        cout << hp.hlpp() << '
    ';
        return 0;
    }
    

    网络流—dinic

    参考题目:https://www.luogu.com.cn/problem/P3376

    #pragma GCC optimize(3,"Ofast","inline")
    #include<bits/stdc++.h>
    #define ios std::ios::sync_with_stdio(false)
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,n,a) for (int i=n;i>=a;i--)
    #define int long long
    #define ll long long
    using namespace std;
    const ll INF (0x3f3f3f3f3f3f3f3fll);
    const int inf = 0x3f3f3f3f3f;
    const int N = 3e5 + 10;
    struct Edge{
        int nex , to;
    }edge[N << 1];
    int head[N] , TOT;
    void add_edge(int u , int v)
    {
        edge[++ TOT].nex = head[u] ;
        edge[TOT].to = v;
        head[u] = TOT;
    }
    const int Max = 1e5 + 10;
    struct Dinic
    {
        struct Edge
        {
            int to, nex, flow;    //flow记录这条边当前的边残量
        } edge[Max << 1];
        int n, m, s, t;
        int head[Max], TOT;
        int dis[Max] , cur[Max];
        void init(int nn , int ss , int tt)
        {
            this->n = nn , this->s = ss , this->t = tt;
            TOT = 1;
            for(int i = 0 ; i <= nn + 1 ; i ++) head[i] = 0;    
        }
        void add_edge(int u , int v , int flow)
        {
            edge[++ TOT].nex = head[u];
            edge[TOT].to = v;
            edge[TOT].flow = flow;
            head[u] = TOT;
            edge[++ TOT].nex = head[v];
            edge[TOT].to = u;
            edge[TOT].flow = 0;
            head[v] = TOT;
        }
        bool bfs() //判断图是否连通
        {
            queue<int>que;
            memset(dis, -1, (max(n + 5 , t + 5)) * sizeof(int));
            dis[s] = 0;
            que.push(s);
            while(!que.empty())
            {
                int u = que.front();
                que.pop();
                for(int i = head[u]; i; i = edge[i].nex)
                {
                    int v = edge[i].to;
                    if(dis[v] == -1 && edge[i].flow > 0)        //可以借助边i到达新的结点
                    {
                        dis[v] = dis[u] + 1;                    //求顶点到源点的距离编号
                        que.push(v);
                    }
                }
            }
            return dis[t] != -1;                                //确认是否连通
        }
        int dfs(int u, int flow_in)
        {
            if(u == t) return flow_in;
            int flow_out = 0;                                    //记录这一点实际流出的流量
            for(int i = cur[u]; i; i = edge[i].nex)
            {
                cur[u] = i;                                     //由于我们增广的时候都是尽量用尽这条边的容量为前提的,
                // 那么在在用尽流入流量之前,我们使用的边,都已经满流了
                //没有继续增广的可能了,所以抛弃了这些重复的边,从未满流的边开始遍历
                int v = edge[i].to;
                if(dis[v] == dis[u] + 1 && edge[i].flow > 0)
                {
                    int flow_part = dfs(v, min(flow_in, edge[i].flow));
                    if(!flow_part)continue;                //无法形成增广路
                    flow_in -= flow_part;                        //流出了一部分,剩余可分配流入就减少了
                    flow_out += flow_part;                        //记录这一点最大的流出
                    edge[i].flow -= flow_part;
                    edge[i ^ 1].flow += flow_part;                //减少增广路上边的容量,增加其反向边的容量
                    if(!flow_in) break;
                }
            }
            return flow_out;
        }
        int dinic()
        {
            int maxflow = 0;
            while(bfs())
            {
                for(int i = 0; i <= n ; i ++) cur[i] = head[i];
                maxflow += dfs(s, inf);
            }
            return maxflow;
        }
    }DI;
    signed main()
    {
        ios;
        cin.tie(0) , cout.tie(0);
        int n , m , s , t;
        cin >> n >> m >> s >> t;
        DI.init(n , s , t);
        rep(i , 1 , m)
        {
            int u , v , w;
            cin >> u >> v >> w;
            DI.add_edge(u , v , w);
        }
        cout << DI.dinic() << '
    ';
        return 0;
    }
    

    字符串哈希

    参考题目:https://www.luogu.com.cn/problem/P3370

    #include<bits/stdc++.h>
    #define ull unsigned long long
    using namespace std;
    const int N = 2e5 + 10;
    const int Max = 2e6 + 10;
    const ull base = 131;
    struct Hash{
    	ull power[Max] , hash1[Max] , hash2[Max];
    	ull get_hash1(int l , int r){
    		return hash1[r] - hash1[l - 1] * power[r - l + 1];
    	}
    	ull get_hash2(int l , int r){
    		return hash2[l] - hash2[r + 1] * power[r - l + 1];
    	}
    	void hash_init(int n)
    	{
    		power[1] = base;
    		for(int i = 2 ; i <= n ; i ++) power[i] = power[i - 1] * base;
    	}
    	void hash_pre(char *s)
    	{
    		int len = strlen(s + 1);
    		hash1[1] = s[1] - 'a' + 1 , hash1[len + 1] = '';
    		for(int i = 2 ; i <= len ; i ++) hash1[i] = hash1[i - 1] * base + s[i] - 'a' + 1;
    	}
    	void hash_suf(char *s)
    	{
    		int len = strlen(s + 1);
    		hash2[len] = s[len] - 'a' + 1 , hash2[len + 1] = '';
    		for(int i = len - 1 ; i >= 1 ; i --) hash2[i] = hash2[i + 1] * base + s[i] - 'a' + 1;
    	}
    }ha;
    char s[N];
    unordered_map<int , int>mp;
    signed main()
    {
    	ha.hash_init(N - 10);
    	int n , res = 0;
    	cin >> n;
    	while(n --)
    	{
    		cin >> s + 1;
    		int len = strlen(s + 1);
    		ha.hash_pre(s);
    		int now = ha.get_hash1(1 , len);
    		if(!mp.count(now)) res ++ , mp[now] ++ ; 
    	}
    	cout << res << '
    ';
    	return 0;
    }
    

    可持久化01trie

    参考题目:https://www.luogu.com.cn/problem/P4735

    #include<bits/stdc++.h>
    #define rep(i , a , b) for(int i = a ; i <= b ; i ++)
    #define per(i , b , a) for(int i = b ; i >= a ; i --)
    using namespace std;
    const int N = 1e6 + 10;
    int n , m , tot;
    int a[N] , sum[N] , cnt[N * 31] , root[N] , tree[N * 31][2];
    void insert(int id , int s)
    {
    	root[id] = ++ tot;
    	int p = root[id] , q = root[id - 1];
    	per(i , 31 , 0)
    	{
    		int x =	s >> i & 1;
    		tree[p][x] = ++ tot;
    		tree[p][x ^ 1] = tree[q][x ^ 1];
    		p = tree[p][x] , q = tree[q][x];
    		cnt[p] = cnt[q] + 1;
    	} 
    }
    int query(int l , int r , int s)
    {
    	int res = 0 , p = root[r] , q = root[l - 1];
    	per(i , 31 , 0)
    	{
    		int x = s >> i & 1;
    		if(cnt[tree[p][x ^ 1]] > cnt[tree[q][x ^ 1]]) 
    		{
    			res += 1LL << i;
    			p = tree[p][x ^ 1] , q = tree[q][x ^ 1];
    		}
    		else p = tree[p][x] , q = tree[q][x];
    	}
    	return res;
    }
    signed main()
    {
    	ios::sync_with_stdio(false);
    	cin.tie(0) , cout.tie(0);
    	cin >> n >> m ;
    	rep(i , 1 , n) cin >> a[i] , sum[i] = sum[i - 1] ^ a[i] , insert(i , sum[i]);
    	while(m --)
    	{
    		char op;
    		cin >> op;
    		if(op == 'A')
    		{
    			cin >> a[++ n];
    			sum[n] = sum[n - 1] ^ a[n];
    			insert(n , sum[n]);
    		}
    		else 
    		{
    			int l , r , x;
    			cin >> l >> r >> x;
    			if(r == 1) 
    			{
    				cout << (sum[n] ^ x) << '
    ';
    				continue ;
    			}
    			cout << query(l - 1 , r - 1 , sum[n] ^ x) << '
    ';
    		}
    	}
    	return 0;
    }
    

    线性求阶层

    #include<bits/stdc++.h>
    #define int long long
    using namespace std;
    const int N = 3e5 + 10 , mod = 998244353;
    int fac[N] , inv[N];
    int pow_mod(int x , int n , int mod)
    {
    	int res = 1;
    	while(n)
    	{
    		if(n & 1) res = res * x % mod;
    		x = x * x % mod;
    		n >>= 1;
    	}
    	return res;
    }
    void init()
    {
    	int up = N - 10;
    	fac[0] = 1;
    	for(int i = 1 ; i <= up ; i ++) fac[i] = fac[i - 1] * i % mod;
    	inv[up] = pow_mod(fac[up] , mod - 2 , mod);
    	for(int i = up - 1 ; i >= 0 ; i --) inv[i] = inv[i + 1] * (i + 1) % mod;
    }
    int C(int n , int m)
    {
    	if(n < m) return 0;
    	return fac[n] * inv[m] % mod * inv[n - m] % mod;
    }
    signed main()
    {
    	init();
    	int n , m ;
    	cin >> n >> m ;
    	cout << C(n , m) << '
    ';
    	return 0;
    }
    
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