【Tarjan】BZOJ1051 [HAOI2006]受欢迎的牛

题目在这里

很显然，缩点后无出度的强连通分量就是答案

示例程序：

#include<cstdio>#include<cstring>#include<algorithm>#define cl(x,y) memset(x,y,sizeof(x))using namespace std;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int red(){    int res=0,f=1;char ch=nc();    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}    while ('0'<=ch&&ch<='9') res=res*10+ch-48,ch=nc();    return res*f;}const int maxn=10005,maxe=50005;int n,e;int tot,son[maxe],nxt[maxe],lnk[maxn];inline void add(int x,int y){    son[++tot]=y;nxt[tot]=lnk[x];lnk[x]=tot;}int stk[maxn],dfn[maxn],low[maxn],times,num[maxn],to[maxn],f[maxn];bool instk[maxn],vis[maxn];void tarjan(int x){    low[x]=dfn[x]=++times;    stk[++stk[0]]=x;instk[x]=1;    for (int j=lnk[x];j;j=nxt[j])     if (!dfn[son[j]]) tarjan(son[j]),low[x]=min(low[x],low[son[j]]);else     if (instk[son[j]]) low[x]=min(low[x],dfn[son[j]]);    if (low[x]==dfn[x]){        while (stk[stk[0]+1]!=x)         num[x]++,to[stk[stk[0]]]=x,         instk[stk[stk[0]--]]=0;    }}int main(){    n=red(),e=red();    for (int i=1,x,y;i<=e;i++) x=red(),y=red(),add(x,y);    for (int i=1;i<=n;i++)     if (!dfn[i]) tarjan(i);    for (int i=1;i<=n;i++)     for (int j=lnk[i];j;j=nxt[j])      if (to[i]!=to[son[j]]) f[to[i]]++;    int ans,tal=0;    for (int i=1;i<=n;i++)     if (num[i]&&!f[i]) ans=num[i],tal++;    if (tal>1) printf("0");else printf("%d",ans);    return 0;}