POJ2186（强连通分量分解）

时间：2024-07-17 21:08:02

Popular Cows

Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 35035   Accepted: 14278

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 3

1 2

2 1

2 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity. 

Source

题意：求从其他所有顶点都可以到达的顶点数目。
思路：所求顶点数目即为拓扑序最后的强连通分量中的顶点数目，检查其他点是否都可以到达该强连通分量。
 //2017-08-20

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 #include <vector>

 using namespace std;

 const int N = ;

 vector<int> G[N];//邻接表存图

 vector<int> rG[N];//存反向图

 vector<int> vs;//后序遍历顺序的顶点列表

 bool vis[N];

 int cmp[N];//所属强连通分量的拓扑序

 void add_edge(int u, int v){

     G[u].push_back(v);

     rG[v].push_back(u);

 }

 //input: u 顶点

 //output: vs 后序遍历顺序的顶点列表

 void dfs(int u){

     vis[u] = true;

     for(int i = ; i < G[u].size(); i++){

         int v = G[u][i];

         if(!vis[v])

               dfs(v);

     }

     vs.push_back(u);

 }

 //input: u 顶点编号; k 拓扑序号

 //output: cmp[] 强连通分量拓扑序

 void rdfs(int u, int k){

     vis[u] = true;

     cmp[u] = k;

     for(int i = ; i < rG[u].size(); i++){

         int v = rG[u][i];

         if(!vis[v])

               rdfs(v, k);

     }

 }

 //Strongly Connected Component 强连通分量

 //input: n 顶点个数

 //output: k 强连通分量数;

 int scc(int n){

     memset(vis, , sizeof(vis));

     vs.clear();

     for(int u = ; u < n; u++)

         if(!vis[u])

               dfs(u);

     int k = ;

     memset(vis, , sizeof(vis));

     for(int i = vs.size()-; i >= ; i--)

           if(!vis[vs[i]])

               rdfs(vs[i], k++);

     return k;

 }

 void solve(int n){

     int k = scc(n);

     int u = , ans = ;

     for(int v = ; v < n; v++){

         if(cmp[v] == k-){

             u = v;

             ans++;

         }

     }

     memset(vis, , sizeof(vis));

     rdfs(u, );

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

         if(!vis[i]){

             ans = ;

             break;

         }

     }

     printf("%d\n", ans);

 }

 int main()

 {

     int n, m;

     while(scanf("%d%d", &n, &m)!=EOF){

         int u, v;

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

             G[i].clear();

             rG[i].clear();

         }

         while(m--){

             scanf("%d%d", &u, &v);

             u--; v--;

             add_edge(u, v);

         }

         solve(n);

     }

     return ;

 }

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