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  • 【luogu P2863 [USACO06JAN]牛的舞会The Cow Prom】 题解

    题目链接:https://www.luogu.org/problemnew/show/P2863
    求强连通分量大小>自己单个点的

    #include <stack>
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    const int maxn = 100000 + 10;
    struct edge{
    	int next, to, from, len;
    }e[maxn<<2];
    int head[maxn], cnt;
    int n, m, color[maxn], tong[maxn], num, tim, ans;
    int dfn[maxn], low[maxn];
    bool vis[maxn];
    stack<int> s;
    void add(int u, int v)
    {
    	e[++cnt].from = u;
    	e[cnt].to = v;
    	e[cnt].next = head[u];
    	head[u] = cnt;
    }
    void tarjan(int x)
    {
    	dfn[x] = low[x] = ++tim;
    	s.push(x); vis[x] = 1;
    	for(int i = head[x]; i != -1; i = e[i].next)
    	{
    		int v = e[i].to;
    		if(!dfn[v])
    		{
    			tarjan(v);
    			low[x] = min(low[x], low[v]);
    		}
    		else if(vis[v])
    		{
    			low[x] = min(low[x], low[v]);
    		}
    	}
    	if(dfn[x] == low[x])
    	{
    		color[x] = ++num;
    		vis[x] = 0;
    		while(s.top() != x)
    		{
    			color[s.top()] = num;
    			vis[s.top()] = 0;
    			s.pop(); 
    		}
    		s.pop();
    	}
    }
    int main()
    {
    	memset(head, -1, sizeof(head));
    	scanf("%d%d",&n,&m);
    	for(int i = 1; i <= m; i++)
    	{
    		int u, v;
    		scanf("%d%d",&u,&v);
    		add(u,v);
    	}
    	for(int i = 1; i <= n; i++)
    		if(!dfn[i]) tarjan(i);
    	for(int i = 1; i <= n; i++)
    		tong[color[i]]++;
    	for(int i = 1; i <= n; i++)
    		if(tong[i] > 1) ans++;
    	printf("%d",ans);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/MisakaAzusa/p/9368270.html
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