poj2186 Popular Cows（Tarjan+缩点水）

http://poj.org/problem?id=2186

题意：给你n头牛m个想法，(u,v)代表牛u认为v很受欢迎，欢迎度可以传递，求所有牛欢迎的牛有几只。

思路：直接求叶子节点出度为0的个数，如果为一个，则输出该连通分量中的元素个数。如果为多个，则输出0。因为缩点后变为有向无环图，所以至少有一个叶子节点。

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <vector>

using namespace std;

typedef long long LL;

const int N = 100010;
const int INF = 1e8;

stack<int>S;

int dfn
, low
, head
, Belong
, in
, out
, countt, time, n, m, pos;
bool instack
;

struct Edge
{
int to, next;
}edge
;

void add(int u, int v)
{
edge[pos].to = v;
edge[pos].next = head[u];
head[u] = pos++;
}

void init()
{
time = countt = pos = 0;
memset(in, 0, sizeof(in));
memset(out, 0, sizeof(out));
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
memset(head, -1, sizeof(head));
memset(instack, false, sizeof(instack));
}

void Tarjan(int u)
{
int v;
dfn[u] = low[u] = ++time;
instack[u] = true;
S.push(u);
for(int i = head[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if(dfn[v] == 0)
{
Tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(instack[v] == 1)
{
low[u] = min(low[u], dfn[v]);
}
}
if(dfn[u] == low[u])
{
countt ++;
do
{
v = S.top();
S.pop();
instack[v] = false;
Belong[v] = countt;
}while(u != v);
}
}

int main()
{
// freopen("in.txt", "r", stdin);
int u, v;
scanf("%d%d", &n, &m);
init();
for(int i = 1; i <= m; i++)
{
scanf("%d%d", &u, &v);
add(u, v);
}
//缩点
for(int i = 1; i <= n; i++)
{
if(dfn[i] == 0)
Tarjan(i);
}
for(int i = 1; i <= n; i++)
{
for(int k = head[i]; k != -1; k = edge[k].next)
{
int j = edge[k].to;
if(Belong[i] != Belong[j])
{
out[Belong[i]]++;
in[Belong[j]]++;
}
}
}
//找叶子和根度为0的数量
int leaf = 0, ans = 0, flag;
for(int i = 1; i <= countt; i++)
{
if(out[i] == 0)
{
leaf++;
flag = i;
}
}
if(leaf == 1)
{
for(int i = 1; i <= n; i++)
{
if(Belong[i] == flag) ans++;
}
printf("%d\n", ans);
}
else
{
printf("0\n");
}
return 0;
}