poj 2186_Popular Cows_强连通分量

题目大意

n头奶牛，给出若干个欢迎关系a b，表示a欢迎b，欢迎关系是单向的，但是是可以传递的。另外每个奶牛都是欢迎他自己的。求出被所有的奶牛欢迎的奶牛的数目

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思路

用tarjan求出强连通分量
如果有多个出度为0的强连通分量的话那么就是无解的，因为肯定有两头牛不互相喜欢
如果只有一个强连通分量，那么全部牛都是popular cows
如果只有一个出度为0的强连通分量，那么求出这个强连通分量中的点的个数就是答案

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#include <stdio.h>#include <stack>#include <string>#include <cstring>using namespace std;#define maxn 10001#define maxm 50005#define fill(x, y) memset(x, y, sizeof(y))int bcnt = 0, dindex = 0;int ls[maxm], dfn[maxn], low[maxn], belong[maxn], in[maxn], out[maxn]; bool instack[maxn];int n, m;struct edge{    int from, to ,next;}e[maxm];stack<int> s;int tarjan(int x){    dindex++;    dfn[x] = dindex;    low[x] = dindex;    s.push(x);    instack[x] = 1;    for (int i = ls[x]; i; i = e[i].next)    {        int to = e[i].to;        if (dfn[to] == 0)        {            tarjan(to);            if (low[x] > low[to]) low[x] = low[to];        }        else        {            if (instack[to] && dfn[to] < low[x])                 low[x] = dfn[to];        }    }    if (low[x] == dfn[x])    {        bcnt++;        int j;        do        {            j = s.top();            s.pop();            instack[j] = 0;            belong[j] = bcnt;        }        while (x != j);    }    return 0;}int solve(){    while (! s.empty()) s.pop();    fill(dfn, 0);    for (int i = 1; i <= n; i++)        if (dfn[i] == 0) tarjan(i);    return 0;}int main(){    while(~scanf("%d%d", &n, &m))    {            int l = 0;    for (int i = 1; i <= m; i++)    {        int x, y;        scanf("%d%d", &x, &y);        e[++l].from = x;        e[l].to = y;        e[l].next = ls[x];        ls[x] = l;    }    solve();    if (bcnt == 1)     {        printf("%d\n", n);        continue;    }    for (int i = 1; i <= m; i++)    {        if (belong[e[i].from] != belong[e[i].to])        {            out[belong[e[i].from]]++;            in[belong[e[i].to]]++;        }    }    int ans = 0, u = 0;    for (int i = 1; i <= bcnt; i++)    {        if (out[i] == 0)         {            ans++;            u = i;        }    }    if (ans > 1) printf("%d\n", 0);    else    {        int an = 0;        for (int i = 1; i <= n; i++)            if (belong[i] == u) an++;        printf("%d\n", an);    }    }}