pku 2186

//思路：强连通分支然后重新构图，然后再判断出度为0的强连通分支里面的元素，如果出度为0的强连通有多个，则一定没有解！！

#include <iostream>#include <string>#include <vector>#define maxn 10001using namespace std;int n,m,vis[maxn],ord[maxn],cnt,degree[maxn],out[maxn],ans;vector<int>G[maxn];vector<int>G1[maxn];void dfs(int u){int i;vis[u] = 1;for(i = 0; i<G[u].size(); i++)if(!vis[G[u][i]])dfs(G[u][i]);ord[cnt++] = u;}void dfs1(int u){int i;degree[u] = ans;vis[u] = 1;for(i = 0; i<G1[u].size(); i++)if(!vis[G1[u][i]])dfs1(G1[u][i]);}int main(){int i,u,v,j,p,sum,f;while(scanf("%d %d",&n,&m)!=EOF){for(i = 1; i<=n; i++){G[i].clear();G1[i].clear();}for(i = 0; i<m; i++){scanf("%d %d",&u,&v);G[u].push_back(v);G1[v].push_back(u);}memset(out,0,sizeof(out));memset(vis,0,sizeof(vis));sum = p = ans = cnt = 0;for(i = 1; i<=n; i++)if(!vis[i])dfs(i);memset(vis,0,sizeof(vis));for(i = cnt - 1; i>=0; i--)if(!vis[ord[i]]){ans++;dfs1(ord[i]);}for(i = 1; i<=n; i++)for(j = 0; j<G[i].size(); j++)//找出度为0的点；{if(degree[i]!=degree[G[i][j]]){out[degree[i]]++;}}for(i = 1; i<=ans; i++)//标志出度为0的点，如果有多个点，此图肯定不连通if(out[i] == 0){p++;f = i;}if(p>1)//不连通{printf("%d/n",0);continue;}for(i = 1; i<=n; i++)//度数为f强连通里面的数if(degree[i] == f)sum++;printf("%d/n",sum);}return 0;}