【USACO】The Cow Prom

【题目链接】

          点击打开链接

【算法】

        tarjan求强连通分量

【代码】

        

#include<bits/stdc++.h>
#define MAXN 20005

using namespace std; 

int i,N,M,u,v,id,cnt;
int dfn[MAXN],low[MAXN],vis[MAXN];
vector<int> edge[MAXN];
stack<int> stk;

void dfs(int u) {
        int i,to,sum;
        dfn[u] = low[u] = ++id;
        stk.push(u); vis[u] = 1;
        for (i = 0; i < edge[u].size(); i++) {
                to = edge[u][i];
                if (!dfn[to]) {
                        dfs(to);
                        low[u] = min(low[u],low[to]);
                }    
                else {
                        if (vis[to])
                                low[u] = min(low[u],dfn[to]);
                }
        }    
        if (dfn[u] == low[u]) {
                sum = 0;
                while (true) {
                        sum++;
                        vis[stk.top()] = 0;
                        if (stk.top() == u) {
                                stk.pop();
                                break;
                        }
                        stk.pop();
                }
                cnt += (sum > 1);
        }
}

int main() {
    
        scanf("%d%d",&N,&M);
        for (i = 1; i <= M; i++) {
                scanf("%d%d",&u,&v);
                edge[u].push_back(v);        
        }
        
        for (i = 1; i <= N; i++) {
                if (!dfn[i])
                        dfs(i);
        }
        
        printf("%d
",cnt);
        
        return 0;
    
}
原文地址:https://www.cnblogs.com/evenbao/p/9196429.html