POJ2186 Popular Cows 强连通分量

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题目链接：http://poj.org/problem?id=2186

今天复习一下图论强连通，到poj上找到裸题，刷着玩一下，顺便水篇博客。

大概就是强连通缩一下点，然后我们看一下，缩点后的点如果有多个出度为0，则不存在，否则的话唯一一个出度为零的即为所求。

#include <cstdio>#include <algorithm>#include <stack>using namespace std;stack <int> dl;const int MAXN = 150000;int head[MAXN],to[MAXN],nxt[MAXN],dfn[MAXN],low[MAXN],ins[MAXN],sg[MAXN];int oud[MAXN],sum[MAXN];int cnt,n,m,a,b,tot,nfg,tjs;void ad_edg(int x,int y){nxt[++tjs] = head[x];head[x] = tjs;to[tjs] = y;}void sread(){scanf("%d%d",&n,&m);for (int i = 1;i <= m;i++){scanf("%d%d",&a,&b);if (a == b) continue;ad_edg(a,b);}}void tarjan(int x){dfn[x] = low[x] = ++cnt;dl.push(x),ins[x] = 1;for (int i = head[x];i;i = nxt[i]){if (!dfn[to[i]]){tarjan(to[i]);low[x] = min(low[x],low[to[i]]);}else if (ins[to[i]])low[x] = min(low[x],dfn[to[i]]);}if (low[x] == dfn[x]){sg[x] = ++tot,sum[tot]++;while (dl.top() != x) ins[dl.top()] = 0,sg[dl.top()] = tot,dl.pop(),sum[tot]++;ins[x] = 0,dl.pop();}}void swork(){for (int i = 1;i <= n;i++)if (!dfn[i]) tarjan(i);for (int i = 1;i <= n;i++)for (int j = head[i];j;j = nxt[j])if (sg[i] != sg[to[j]])oud[sg[i]]++;cnt = 0,a = 0;for (int i = 1;i <= tot;i++)if (!oud[i])cnt++,a = i;if(cnt > 1 || !cnt) printf("0\n");elseprintf("%d\n",sum[a]);}int main(){sread();swork();return 0;}

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