poj2186Popular Cows【scc+缩点】

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Popular Cows

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 27987 Accepted: 11275

Description

Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is 
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow. 

Input

* Line 1: Two space-separated integers, N and M 

* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular. 

Output

* Line 1: A single integer that is the number of cows who are considered popular by every other cow. 

Sample Input

3 31 22 12 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity. 

Source

USACO 2003 Fall

题意：没头牛都有认为一头牛最知名具有传递性求被所有牛都认为知名的牛的个数

解题思路：scc+缩点如果为强连通图数目为n缩点后出度大于1则不存在；出度为一时输出出度为1的scc的牛数即可

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<stack>#include<vector>#define inf 0x3f3f3f3f3fusing namespace std;const int maxn=100010;int dfs_clock,scc_cnt;int head[maxn];bool instack[maxn];int low[maxn],dfn[maxn];int sccno[maxn];int in[maxn],out[maxn];stack<int>S;vector<int>G[maxn];vector<int>scc[maxn];struct Node{int from,to,next;}A[maxn];void init(){dfs_clock=scc_cnt=0;memset(head,-1,sizeof(head));memset(low,0,sizeof(low));memset(dfn,0,sizeof(dfn));memset(in,0,sizeof(in));memset(out,0,sizeof(out));memset(sccno,0,sizeof(sccno));memset(instack,false,sizeof(instack));}int MIN(int a,int b){return a<b?a:b;}void tarjan(int u,int pre){low[u]=dfn[u]=++dfs_clock;S.push(u);instack[u]=true;int v;for(int k=head[u];k!=-1;k=A[k].next){v=A[k].to;if(!dfn[v]){tarjan(v,u);low[u]=MIN(low[u],low[v]);}else if(instack[v]){low[u]=MIN(low[u],dfn[v]);}}if(low[u]==dfn[u]){scc_cnt++;G[scc_cnt].clear();scc[scc_cnt].clear();while(1){v=S.top();S.pop();instack[v]=false;sccno[v]=scc_cnt;scc[scc_cnt].push_back(v);if(u==v)break;}}}void suodian(int m){for(int i=0;i<m;++i){int u=sccno[A[i].from];int v=sccno[A[i].to];if(u!=v){G[u].push_back(v);in[v]++;out[u]++;}}}int main(){int n,m,i,j,k;while(scanf("%d%d",&n,&m)!=EOF){init();for(i=0;i<m;++i){scanf("%d%d",&A[i].from,&A[i].to);A[i].next=head[A[i].from];head[A[i].from]=i;}for(i=1;i<=n;++i){if(!dfn[i])tarjan(i,-1);}suodian(m);int ans=0,pos;for(i=1;i<=scc_cnt;++i){if(!out[i]){ans++;pos=i;}}if(scc_cnt==1){printf("%d\n",n);}else if(ans>1){printf("%d\n",0);}else printf("%d\n",scc[pos].size());}return 0;}