不裸缩点》。。。POJ2186受欢迎的牛

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1051: [HAOI2006]受欢迎的牛

Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 4930 Solved: 2626
[Submit][Status][Discuss]

Description

　　每一头牛的愿望就是变成一头最受欢迎的牛。现在有N头牛，给你M对整数(A,B)，表示牛A认为牛B受欢迎。这

种关系是具有传递性的，如果A认为B受欢迎，B认为C受欢迎，那么牛A也认为牛C受欢迎。你的任务是求出有多少头

牛被所有的牛认为是受欢迎的。

Input

　　第一行两个数N,M。接下来M行，每行两个数A,B，意思是A认为B是受欢迎的（给出的信息有可能重复，即有可

能出现多个A,B）

Output

　　一个数，即有多少头牛被所有的牛认为是受欢迎的。

Sample Input

3 3
1 2
2 1
2 3

Sample Output

HINT

100%的数据N<=10000,M<=50000

来源： http://www.lydsy.com:808/JudgeOnline/problem.php?id=1051

#include<cstdio>

#include<cstring>

#include<iostream>

#include<string>

#include<vector>

#include<stack>

#include<algorithm>

using namespace std;

#define N 10002

vector<int>Gra[N];

stack<int>Sta;

int low[N],dfn[N],inStack[N],belong[N],sum[N],outDegree[N],Time,cnt;

void Tarjan(int s)

{

dfn[s] = low[s] = ++Time;

Sta.push(s); inStack[s] = 1;

for(int i=0;i<Gra[s].size();i++)

{

int j = Gra[s][i];

if(dfn[j] == 0){

Tarjan(j);

low[s] = min(low[s], low[j]);

}

else if(inStack[j]){

low[s] = min(low[s], dfn[j]);

}

if(dfn[s] == low[s])

{

cnt++;

while(!Sta.empty()){

int k = Sta.top(); Sta.pop();

sum[cnt] ++;

belong[k] = cnt;

inStack[s] = 0;

if(k == s) break;

}

return ;

}

int main()

{

int n,m,x,y;

scanf("%d%d",&n,&m);

for(int i=0;i<m;i++)

{

scanf("%d%d",&x,&y);

Gra[x].push_back(y);

}

Time = cnt = 0;

memset(sum,0,sizeof(sum));

memset(inStack,0,sizeof(inStack));

memset(outDegree,0,sizeof(outDegree));

for(int i=1;i<=n;i++) if(dfn[i] == 0) Tarjan(i);

for(int i=1;i<=n;i++)

{

for(int j=0;j<Gra[i].size();j++){

int k = Gra[i][j];

if(belong[i] != belong[k]){

outDegree[belong[i]]++;

}

int outDegree_0 = 0;

int ret ;

for(int i=1;i<=cnt;i++)

{

if(outDegree[i] == 0){

outDegree_0 ++;

ret = sum[i];

}

if(outDegree_0 > 1) printf("0");

else printf("%d",ret);

}

来源： http://tool.oschina.net/highlight