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【Edu 67】补题记录

CF1187D. Subarray Sorting

想要把一个数x换到前面，x一定是小一点的值。由于B串是固定的，A串可调整，我们可以遍历B数组，对于B【i】，找到对于在A数组的位子pos，判断1～pos中，是不是A【pos】最小，如果是最小，说明可以换到最前面，把A【pos】的值变为inf后，继续遍历。

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
 
const int inf = 0x3f3f3f3f;
 
const int MOD = 1e9+7;
 
/**********showtime************/
 
            const int maxn = 3e5+9;
            int a[maxn], b[maxn];
            queue<int>que[maxn];
            int mn[maxn<<2];
            void update(int pos, int val, int le, int ri, int rt) {
                if(le == ri) {
                    mn[rt] = val;
                    return ;
                }
                int mid = (le + ri) >> 1;
                if(mid >= pos) update(pos, val, le, mid, rt<<1);
                else update(pos, val, mid+1, ri, rt<<1|1);
                mn[rt] = min(mn[rt<<1], mn[rt<<1|1]);
            }
            int query(int L, int R, int le, int ri, int rt) {
                if(le >= L && ri <= R) {
                    return mn[rt];
                }
                int mid = (le + ri) >> 1;
                int res = inf;
                if(mid >= L) res = min(res, query(L,R,le, mid, rt<<1));
                if(mid < R) res = min(res, query(L,R,mid+1, ri, rt<<1|1));
                return res;
            }
            void build(int le, int ri, int rt) {
                if(le == ri) return ;
                int mid = (le + ri) >> 1;
                build(le, mid, rt<<1);
                build(mid+1, ri, rt<<1|1);
            }
int main(){
            int T;  scanf("%d", &T);
            while(T--) {
                int n;  scanf("%d", &n);
                build(1, n, 1);
                for(int i=1; i<=n; i++) scanf("%d", &a[i]), que[a[i]].push(i), update(i, a[i], 1, n, 1);
                for(int i=1; i<=n; i++) scanf("%d", &b[i]);
                int flag = 1;
                for(int i=1; i<=n; i++) {
                    if(!que[b[i]].empty()) {
                        int p = que[b[i]].front();  que[b[i]].pop();
                        if(query(1, p, 1, n, 1) < b[i]) flag = 0;
                        else update(p, inf, 1, n, 1);
                    }
                    else flag = 0;
                }
                for(int i=1; i<=n; i++) {
                    while(!que[a[i]].empty()) que[a[i]].pop();
                }
                puts(flag?"YES":"NO");
            }
            return 0;
}

View Code

E Tree Painting

树形DP，换根的转移。

#include <bits/stdc++.h>
 
using namespace std;
 
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
 
 
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int inf = 0x7f7f7f7f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const double eps = 1e-6;
/**********showtime************/
            const int maxn = 2e5+9;
            vector<int> mp[maxn];
            int sz[maxn];
            ll ans = 0;
            int n;
            void dfs(int u, int fa) {
                sz[u] = 1;
                for(int i=0; i<mp[u].size(); i++) {
                    int v = mp[u][i];
                    if(v == fa) continue;
                    dfs(v, u);
                    sz[u] += sz[v];
                }
                ans += sz[u];
            }
            void dfs2(int u, int fa, ll pre) {
 
                for(int i=0; i<mp[u].size(); i++) {
                    int v = mp[u][i];
                    if(v == fa) continue;
 
                    ll tmp = pre + n - 2ll * sz[v];
                    ans = max(ans, tmp);
 
                    dfs2(v, u, tmp);
                }
            }
int main(){
            scanf("%d", &n);
            for(int i=1; i<n; i++) {
                int u,v;
                scanf("%d%d", &u, &v);
                mp[u].pb(v);
                mp[v].pb(u);
            }
            dfs(1, -1);
            dfs2(1, -1, ans);
            printf("%lld
", ans);
            return 0;
}

View Code

G Gang Up

网络流，搞好了。

这道题要把每个点拆成多个点表示不同的时间，每条边要对应多条道路，就是把1,4,9....变成容量为1,费用位1,3,5的边。

每个点要拆成100个点，就是最多会等待100的时间单位。因为第一个人最迟50个时间，那个第50个人就要100个时间到。

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int inf = 0x3f3f3f3f;

const int mod = 1e9+7;

/**********showtime************/
            const int maxn = 55;
            int st[maxn];

            struct E{
                int u,v,w,cost;
                int nxt;
            } edge[500009];
            int gtot = 0,head[maxn * 100];
            void addedge(int u, int v, int w, int cost) {
                edge[gtot].v = v;
                edge[gtot].w = w;
                edge[gtot].cost = cost;
                edge[gtot].nxt = head[u];
                head[u] = gtot++;


                edge[gtot].v = u;
                edge[gtot].w = 0;
                edge[gtot].cost = -cost;
                edge[gtot].nxt = head[v];
                head[v] = gtot++;
            }
            int dis[maxn * 100],path[500009],vis[maxn*100],pre[maxn*100];
            bool spfa(int s,int t){
                memset(pre, -1 ,sizeof(pre));
                memset(dis, inf,sizeof(dis));
                memset(vis, 0, sizeof(vis));

                queue<int>que;  que.push(s);
                vis[s] = 1; dis[s] = 0;
                while(!que.empty()){
                    int u = que.front(); que.pop();
                    vis[u] = 0;
                    for(int i=head[u]; ~i; i = edge[i].nxt){
                        int v = edge[i].v, val = edge[i].w, cost = edge[i].cost;

                        if(val > 0 && dis[v] > dis[u] + cost){
                            dis[v] = dis[u] + cost;
                            pre[v] = u; path[v] = i;
                            if(vis[v] == 0){
                                vis[v] = 1;
                                que.push(v);
                            }
                        }
                    }
                }
                return pre[t] != -1;
            }
            int mcmf(int s,int t){
                int flow = 0,cost = 0;

                while(spfa(s, t)){
                    int f = inf;
                    for(int i=t; i!=s; i = pre[i]){
                        f = min(f, edge[path[i]].w);
                    }
                    flow += f;
                    cost += f * dis[t];
                    for(int i=t; i!=s; i = pre[i]){
                        edge[path[i]].w -= f;
                        edge[path[i]^1].w += f;
                    }
                }
                return cost;
            }
int main(){
            memset(head, -1, sizeof(head));
            gtot = 0;
            int n,m,k,c,d;
            scanf("%d%d%d%d%d", &n, &m, &k, &c, &d);
            for(int i=1; i<=k; i++) {
                int x;
                scanf("%d", &x);
                st[x]++;
            }

            int s = 0, t = 100 * n + 1;

            for(int i=1; i<=n; i++) {
                for(int j=2; j<=100; j++) {
                    addedge((i-1)*100+j-1, (i-1)*100 + j, inf, c);
                }
            }

            for(int i=2; i<=n; i++) {
                if(st[i]) addedge(s, (i-1)*100 + 1, st[i], 0);
            }

            for(int i=1; i<=100; i++) addedge(i, t, inf, 0);

            for(int i=1; i<=m; i++) {
                int u,v;
                scanf("%d%d", &u, &v);
                for(int j=2; j<=100; j++) {
                    for(int k=1; k<=20; k++){
                        addedge((u-1)*100 + j - 1, (v-1)*100 + j, 1, (k*k - (k-1)*(k-1)) * d + c);
                        addedge((v-1)*100 + j - 1, (u-1)*100 + j, 1, (k*k - (k-1)*(k-1)) * d + c);
                    }
                }
            }

            printf("%d
", mcmf(s, t));
            return 0;
}

View Code

也可以两个点之间只建立一条道路，那么费用就需要根据流量重新计算。

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
 
 
template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(ll &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }
 
const int inf = 0x3f3f3f3f;
 
const int mod = 1e9+7;
 
/**********showtime************/
            const int maxn = 55;
            int st[maxn];
 
            // Min-Cost Flow with positive convex edge weights
class MinCostFlow {
    private:
        struct Edge {
            int a, b; // source, target
            int f, c; // flow, capacity
            ll m1, m2; // constants for cost
 
            Edge(int av, int bv, int cv, ll m1v, ll m2v)
                : a(av), b(bv), f(0), c(cv), m1(m1v), m2(m2v) {}
 
            // Get other end of edge
            int getOth(int i) {
                return i == a ? b : a;
            }
            // Get change in cost of pushing flow from i to getOth(i)
            ll getCost(int i) {
                int tf = f + (i == a ? 1 : -1);
                if (tf < 0 || tf > c) return inf;
                return m1*(tf-f) + m2*(tf*tf-f*f);
            }
            void pushUnit(int i) {
                f += (i == a ? 1 : -1);
            }
        };
 
        vector<Edge> eds;
        vector<vector<int> > conns;
    public:
        MinCostFlow(int n) : conns(n) {}
 
        void addEdge(int a, int b, int c, int m1, int m2) {
            Edge tmp(a, b, c, m1, m2);
            eds.push_back(tmp);
            conns[a].push_back(eds.size() - 1);
            conns[b].push_back(eds.size() - 1);
        }
        ll pushFlow(int source, int sink) {
            int n = conns.size();
            ll res = 0;
 
            vector<int> pre(n);
            vector<ll> cost(n);
            while(true) {
                for (int i = 0; i < n; ++i) {
                    pre[i] = -1;
                    cost[i] = inf;
                }
                cost[source] = 0;
 
                priority_queue<pair<ll, int> > que;
                que.push(pair<ll, int>(0, source));
                while(! que.empty()) {
                    int i = que.top().second;
                    ll v = -que.top().first;
                    que.pop();
                    if (v > cost[i]) continue;
                    for (auto j : conns[i]) {
                        int t = eds[j].getOth(i);
                        ll off = cost[i] + eds[j].getCost(i);
                        if (off < cost[t]) {
                            cost[t] = off;
                            pre[t] = j;
                            que.push(pair<ll, int>(-off, t));
                        }
                    }
                }
                int i = sink;
                while(pre[i] != -1) {
                    auto& ed = eds[pre[i]];
                    i = ed.getOth(i);
                    ed.pushUnit(i);
                }
                if (i != source) break;
                res += cost[sink];
            }
            return res;
        }
};
int main(){
            int n,m,k,c,d;
          //  scanf("%d%d%d%d%d", &n, &m, &k, &c, &d);
            R(n,m,k,c,d);
            
            for(int i=1; i<=k; i++) {
                int x;
                scanf("%d", &x);
                st[x]++;
            }
            
            int s = 0, t = 100 * n + 1;
            MinCostFlow fl(t + 1);
            
            for(int i=1; i<=n; i++) {
                for(int j=2; j<=100; j++) {
                    fl.addEdge((i-1)*100+j-1, (i-1)*100 + j, inf, c, 0);
                }
            }
 
            for(int i=2; i<=n; i++) {
                if(st[i]) fl.addEdge(s, (i-1)*100 + 1, st[i], 0, 0);
            }
 
            for(int i=1; i<=100; i++) fl.addEdge(i, t, inf, 0, 0);
 
            for(int i=1; i<=m; i++) {
                int u,v;
                scanf("%d%d", &u, &v);
                for(int j=2; j<=100; j++) {
                        fl.addEdge((u-1)*100 + j - 1, (v-1)*100 + j, k, c, d);
                        fl.addEdge((v-1)*100 + j - 1, (u-1)*100 + j, k, c, d);
                }
            }
 
            // printf("%d
", fl.mcmf(s, t));
            cout<<fl.pushFlow(s, t)<<endl;
            return 0;
}

View Code

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原文地址：https://www.cnblogs.com/ckxkexing/p/11117214.html

【Edu 67】 补题记录

CF1187D. Subarray Sorting

E Tree Painting

G Gang Up

【Edu 67】补题记录