题目链接

https://ac.nowcoder.com/acm/problem/51319

题意

求每个男孩可以选择多少个女孩使得完美匹配依然存在。

思路

建图时， 男孩对喜欢的女孩建一条单向边， 和暂时匹配的女孩建一条双向边， 当男孩和女孩同处于一个强连通分量并且图中有边， 男孩就可以选择这个女孩。

AC代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5000;
const int INF = 0x3f3f3f3f;
struct node{
    int to, next;
} edge[maxn * 100];
int tot, sc, cnt;
int dfn[maxn], low[maxn];
int top, stack1[maxn * 100];
int scc[maxn];
int head[maxn], vis[maxn];
int sz[maxn];
void add(int from, int to){
    edge[++cnt].to = to;edge[cnt].next = head[from];head[from] = cnt;
}
void ins(int u, int v){add(u, v); add(v, u);}
void Tarjan(int u){
    dfn[u] = low[u] = ++tot;
    vis[u] = 1;
    stack1[top++] = u;
    for(int i = head[u];i != -1;i = edge[i].next){
        int v = edge[i].to;
        if(!dfn[v]){
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(vis[v]) low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u]){
        sc++;
        int v;
        do{
            v = stack1[--top];
            scc[v] = sc;
            vis[v] = 0;
            sz[sc]++;
        }while(v != u);
    }
}
void init(){
    tot = cnt = sc = 0;
    memset(head, -1, sizeof(head));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(scc, 0, sizeof(scc));
}
int main()
{
    std::ios::sync_with_stdio(false);
    int n;
    while(cin >> n){
        init();
        for(int i = 1;i <= n;i++){
            int m;
            cin >> m;
            for(int j = 0;j < m;j++){
                int x;
                cin >> x;
                add(i, x + n);
            }
        }
        for(int i = 1;i <= n;i++){
            int x;
            cin >> x;
            ins(i, x + n);
        }
        for(int i = 1;i <= 2 * n;i++) if(!dfn[i]) Tarjan(i);
        for(int i = 1;i <= n;i++){
            set<int> ans;
            for(int j = head[i];j != -1;j = edge[j].next){
                int v = edge[j].to;
                if(scc[i] == scc[v]) ans.insert(v - n);
            }
            cout << ans.size();
            for(auto j : ans) cout << " " << j;
            cout << endl;
        }
    }
    return 0;
}