算法竞赛入门经典第二版习题选练

uva839 （递归来判断天平是否平衡）（递归+树）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 2000+10;

bool solve(int& W){
    int w1, l1, w2, l2;
    cin>>w1>>l1>>w2>>l2;
    bool flag1, flag2;
    flag1 = flag2 = true;
    if(w1 == 0) flag1 = solve(w1);
    if(w2 == 0) flag2 = solve(w2);
    W = w1+w2;
    if(flag1&&flag2&&w1*l1 == w2*l2) return true;
    return false;
}


int main(){
    ios::sync_with_stdio(false);
    int _;
    cin>>_;
    int m;
    while(_--){
        if(solve(m)) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
        if(_) cout<<endl;
    }


    return 0;
}

四分树的递归统计 （分治）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 2000+10;
const int len = 32;
char s[1024+10];
int ans = 0;
bool vis[len][len];

void draw(char *s, int &p, int r, int c, int w){
    char ch = s[p++];
    if(ch == 'p'){
        draw(s, p, r, c+w/2, w/2);
        draw(s, p, r, c, w/2);
        draw(s, p, r+w/2, c, w/2);
        draw(s, p, r+w/2, c+w/2, w/2);
    }
    else if(ch == 'f'){

        for(int i=r; i<r+w; i++){
            for(int j=c; j<c+w; j++){
                if(!vis[i][j]) {
                    vis[i][j] = true;
                    ans++;
                }
            }
        }

    }
}

int main(){
    ios::sync_with_stdio(false);
    int _;
    cin>>_;
    while(_--){
        cls(vis, 0);
        ans = 0;
        for(int i=0; i<2; i++){
            cin>>s;
            int p=0;
            draw(s, p, 0, 0, len);
        }
        cout<<"There are "<<ans<<" black pixels."<<endl;
    }

    return 0;
}

四分树2

petri网络的模拟（模拟）

一个P节点有若干的资源，一个T(transaction)节点在入节点有足够多的资源之后，会触发事件，并且使出节点的资源增加。问是否是一个循环的网络，若在规定的时间内出现了资源不足的情况，那么该网络就死了。

问最后每一个节点的最后的资源的个数。

我WA了三发就是因为输出的格式有问题。。。

问题本身是一个相对简单的模拟，按照顺序模拟就可以了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 100+10;
int NP, NT;
int token[maxn];
struct Node{
    bool in[maxn], out[maxn];
    //vector<int> iid, oid;
    int din[maxn], dout[maxn];

}node[maxn];

int main(){
    ios::sync_with_stdio(false);
    int kase=1;
    while(~scanf("%d", &NP)&&NP){

        for(int i=1; i<=NP; i++) scanf("%d", &token[i]);

        scanf("%d", &NT);
        for(int i=1; i<=NT; i++){
            cls(node[i].in, 0), cls(node[i].out, 0);
            //node[i].iid.clear(), node[i].oid.clear();
            cls(node[i].din, 0), cls(node[i].dout, 0);
        }
        for(int i=1; i<=NT; i++){
            int id;
            while(scanf("%d", &id)&&id){
                if(id<0){
                    node[i].in[-id] = true, node[i].din[-id]++;
                }
                else node[i].out[id]=true, node[i].dout[id]++;
            }
        }
        int cnt = 0;
        int lim;
        scanf("%d", &lim);

        for(int i=1; i<=NT; i++){
            bool trig = true;
            for(int j=1; j<=NP; j++){
                if(node[i].in[j]&&node[i].din[j]>token[j]){
                    trig = false;
                    break;
                }
            }
            if(!trig) continue;
            for(int j=1; j<=NP; j++){
                token[j] -= node[i].din[j];
                token[j] += node[i].dout[j];
            }
            i=0;
            cnt++;
            if(cnt>=lim) break;
        }
        if(cnt>=lim){
            printf("Case %d: still live after %d transitions
", kase++, lim);

        }
        else{
            printf("Case %d: dead after %d transitions
", kase++, cnt);
        }
        printf("Places with tokens:");
        for(int i=1;i<=NP;++i) if(token[i]) printf(" %d (%d)",i,token[i]);
        printf("

");
    }

    return 0;
}

列车调度 （栈）

栈的模拟，uva的输入格式真的很奇葩

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 2000+10;
int n;
stack<int> s;
int target[maxn];

int main(){
    ios::sync_with_stdio(false);
    while(cin>>n&&n){
        while(true){
            cin>>target[1];
            if(target[1] == 0) {
                cout<<endl;
                break;
            }
            for(int i=2; i<=n; i++) cin>>target[i];
            while(!s.empty()) s.pop();
            int A, B;
            A = B = 1;
            bool ok = true;
            while(B<=n){
                if(A == target[B]) A++, B++;
                else if(!s.empty()&&target[B] == s.top()) B++, s.pop();
                else if(A<=n) s.push(A), A++;
                else {
                    ok = false;
                    break;
                }
            }
            if(ok) cout<<"Yes"<<endl;
            else cout<<"No"<<endl;
        }
    }

    return 0;
}

矩阵的乘法 （栈）

这道题比较特殊的一点就是任何两个矩阵之间都会有括号，这样就会导致最后的stack里面只有一个结果矩阵。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 2000+10;
struct Matrix{
    int a, b;
}node[maxn];
int n;

int main(){
    ios::sync_with_stdio(false);
    cin>>n;
    string name;
    int x, y;
    for(int i=1; i<=n; i++){
        cin>>name>>x>>y;
        node[name[0]-'A'].a = x;
        node[name[0]-'A'].b = y;
    }
    string op;
    stack<Matrix> s;
    while(cin>>op){
        bool flag = false;
        int ans = 0;
        for(int i=0; i<op.length(); i++){
            if(op[i]>='A'&&op[i]<='Z') s.push(node[op[i]-'A']);
            else if(op[i] ==')'){
                Matrix m2 = s.top();s.pop();
                Matrix m1 = s.top();s.pop();
                if(m1.b!=m2.a){
                    flag = true;
                    break;
                }
                ans += m1.a*m2.a*m2.b;
                s.push(Matrix{m1.a, m2.b});
            }
        }
        if(flag){
            cout<<"error"<<endl;
        }
        else cout<<ans<<endl;
    }

    return 0;
}

flip （规律）

注意数组不是一个排列！，每次都选择数组中的最大的数进行操作。

lrj输入输出混用了。。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include <sstream>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 30+10;
int n, a[maxn];

void flip(int p){
    for(int i=0; i<p-i; i++){
        swap(a[i], a[p-i]);
    }
    cout<<n-p<<" ";
}

int main(){
    ios::sync_with_stdio(false);
    string s;
    while(getline(cin, s)){
        cout<<s<<endl;
        n=0;
        stringstream ss(s);
        while(ss>>a[n]) n++;
        //cout<<n<<endl;
        for(int i=n-1; i>=0; i--){
            int pos = max_element(a, a+i+1)-a;
            //cout<<pos<<endl;
            if(pos == i) continue;
            if(pos>0) flip(pos);//将最大的翻上去
            //将最大的翻到最底
            flip(i);

        }
        cout<<0<<endl;
    }

    return 0;
}

uva1471 （数据结构优化dp）

删去连续的一段，使得剩下的序列中，一段连续的增序列的长度最长。

set的lower_bound()的灵活的运用。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 2e5+10;
struct Node{
    int val,g;
    bool operator < (const Node &b) const{
        return val<b.val;
    }
};
int a[maxn];
int n;
set<Node> s;
int g[maxn], f[maxn];

int main(){
    ios::sync_with_stdio(false);
    int _;
    cin>>_;
    while(_--){
        cin>>n;
        for(int i=1; i<=n; i++){
            cin>>a[i];
        }
        g[1] = 1, f[n] = 1;
        s.clear();
        for(int i=2; i<=n; i++){
            if(a[i-1]<a[i]) g[i] = g[i-1]+1;
            else g[i] = 1;
        }
        for(int i=n-1; i>=1; i--){
            if(a[i+1]>a[i]) f[i] = f[i+1]+1;
            else f[i] = 1;
        }
        s.insert(Node{a[1], g[1]});
        int ans = 1;
        for(int i=2; i<=n; i++){
            Node now = Node{a[i], g[i]};
            set<Node>::iterator it = s. lower_bound(now);
            bool have = true;
            if(it!=s.begin()){
                Node last = *(--it);
                ans = max(ans, last.g+f[i]);
                if(g[i]<=last.g) have=false;
            }
            if(have){
                s.erase(now);
                s.insert(now);
                it = s. upper_bound(now);
                while(it!=s.end()&&it->g<=now.g) s.erase(it++);
            }
        }
        cout<<ans<<endl;

    }

    return 0;
}

理想路径 (最小字典序+bfs)

有n个点m条边的无向图，每条边有一个颜色。

问1->n的最短路的颜色最小字典序。

两遍bfs:

1. 第一遍求出各个点到n节点的最短路，其实可以顺着做，也可以逆着做。
2. 从1节点出发，不断的试探最小的颜色，就可以了

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 2e5+10;
int n, m;
struct Node{
    int v, c;
};
vector<Node> G[maxn];

bool vis[maxn];
int d[maxn];

void rev_bfs(){
    cls(vis, 0);
    cls(d, 0x3f);
    queue<int> q;
    q.push(n);
    vis[n] = true;
    d[n] = 0;
    while(!q.empty()){
        int v = q.front();
        q.pop();
        for(int i=0; i<G[v].size(); i++){
            int u = G[v][i].v;
            if(!vis[u]&&d[u]>d[v]+1){
                vis[u] = true;
                d[u] = d[v]+1;
                q.push(u);
            }
        }
    }
}

vector<int> ans;
void bfs(){
    cls(vis, 0);
    vector<int> next, next2;
    ans.clear();
    next.push_back(1);
    vis[1] = true;
    for(int i=0; i<d[1]; i++){
        int minc = inf;
        for(int j=0; j<next.size(); j++){
            int u = next[j];
            for(int k=0; k<G[u].size(); k++){
                int v = G[u][k].v;
                int c = G[u][k].c;
                if(!vis[v] && d[u] == d[v]+1){
                    minc = min(minc, c);
                }
            }
        }
        ans.push_back(minc);
        next2.clear();
        for(int j=0; j<next.size(); j++){
            int u = next[j];
            for(int k=0; k<G[u].size(); k++){
                int v = G[u][k].v;
                int c = G[u][k].c;
                if(!vis[v] && d[u] == d[v]+1&&c == minc){
                    vis[v] = true;
                    next2.push_back(v);
                }
            }
        }
        next = next2;
    }
    cout<<ans.size()<<endl;
    for(int i=0; i<ans.size()-1; i++){
        cout<<ans[i]<<" ";
    }
    cout<<ans[ans.size()-1]<<endl;
}



int main(){
    ios::sync_with_stdio(false);
    while(cin>>n>>m){
        int u, v, c;
        for(int i=1; i<=n; i++) G[i].clear();
        for(int i=0; i<m; i++){
            cin>>u>>v>>c;
            G[u].push_back(Node{v, c});

            G[v].push_back(Node{u, c});
        }
        rev_bfs();
        bfs();
    }

    return 0;
}

抄书(二分+贪心)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 500+10;
int a[maxn];
int n, m;

bool check(int mid){
    int tot = 0 ;
    int cnt = 0;
    for(int i=n; i>=1; i--){
        if(a[i]>mid){
            return false;
        }
        if(tot+a[i]<=mid){
            tot += a[i];
        }
        else{
            cnt++;
            tot = a[i];
        }
    }
    if(tot>0){
        tot = 0;
        cnt++;
    }
    return cnt<=m;
}

int main(){
    ios::sync_with_stdio(false);
    int _;
    cin>>_;
    while(_--){
        cin>>n>>m;
        for(int i=1; i<=n; i++) cin>>a[i];
        int l=0, r = 1e9;
        int ans = 0;
        while(l<r){
            int mid = l+(r-l)/2;
            if(check(mid)){
                ans = mid;
                r=mid;
            }
            else l=mid+1;
        }
        int sum = 0;
        bool in[maxn];
        cls(in, 0);
        //cout<<ans<<endl;

        for(int i=n; i>=1; i--){
            if(i<=m-1){
                in[i] = true;
                m--;
                continue;
            }
            if(sum +a[i]<=ans){
                sum += a[i];
            }
            else{
                sum = a[i];
                in[i] = true;
                m--;
            }
        }
        for(int i=1; i<n; i++){
            cout<<a[i];
            if(in[i]){
                cout<<" / ";
            }
            else cout<<" ";
        }
        cout<<a[n]<<endl;
    }


    return 0;
}

uva1442 洞窟 （扫两遍）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<bitset>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 1e6+10;
int d[maxn];
int t[maxn];
int h[maxn];
int n;

int main(){
    ios::sync_with_stdio(false);
    int _;
    cin>>_;
    while(_--){
        cin>>n;
        for(int i=1; i<=n; i++)cin>>d[i];
        for(int i=1; i<=n; i++)cin>>t[i];
        int level = t[1];
        for(int i=1; i<=n; i++){
            if(d[i]>level) level = d[i];
            if(level>t[i]) level = t[i];
            h[i] = level;
        }
        int ans = 0;
        level = t[n];
        for(int i=n; i>=1; i--){
            if(d[i]>level) level = d[i];
            if(level>t[i]) level = t[i];
            ans += min(h[i], level)-d[i];
        }
        cout<<ans<<endl;
    }

    return 0;
}

uva 1354 mobile computing (枚举二叉树)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
#define pb(x) push_back(x)
#define cls(x, val) memset(x, val, sizeof(x))
#define fi first
#define se second
#define mp(x, y) make_pair(x, y)
#define inc(i, l, r) for(int i=l; i<=r; i++)
const int inf = 0x3f3f3f3f;
const int maxn = 6+10;
int w[maxn];
int sum[1<<6];
int n;
double R;
struct Node{
    double l, r;
};
vector<Node> node[1<<6];
bool vis[1<<6];

void dfs(int x){
    if(vis[x]) return;
    vis[x] = true;
    bool has_child = false;
    //cout<<x<<endl;
    for(int l=(x-1)&x; l; l=(l-1)&x){
        int r = x^l;
        has_child = true;
        double d1 = 1.0*sum[r]/sum[x];
        double d2 = 1.0*sum[l]/sum[x];
        dfs(l), dfs(r);
        for(int i=0; i<node[l].size(); i++){
            for(int j=0; j<node[r].size(); j++){
                Node temp = Node{0, 0};
                temp.l = max(node[l][i].l+d1, node[r][j].l-d2);
                temp.r = max(node[r][j].r+d2, node[l][i].r-d1);
                //cout<<temp.l<<" "<<temp.r<<endl;
                if(temp.l+temp.r<=R) node[x].push_back(temp);
            }

        }

    }
    if(!has_child) node[x].push_back(Node{0, 0});

}

int main(){
    ios::sync_with_stdio(false);
    int _;
    cin>>_;
    while(_--){
        cin>>R>>n;
        for(int i=0; i<n; i++) cin>>w[i];

        for(int i=1; i<(1<<n); i++){
            sum[i] = 0;
            node[i].clear();
            for(int j=0; j<n; j++) if(i&(1<<j)) sum[i]+=w[j];
        }
        cls(vis, 0);
        dfs((1<<n)-1);
        double ans = -1;
        for(int i=0; i<node[(1<<n)-1].size(); i++){
            ans = max(ans, node[(1<<n)-1][i].l+node[(1<<n)-1][i].r);
        }
        if(ans == -1){
            cout<<-1<<endl;
        }
        else{
            cout<<fixed<<setprecision(11);
            cout<<ans<<endl;
        }
    }

    return 0;
}