POJ 2186 Popular Cows

Popular Cows

POJ - 2186

时限: 2000MS 内存: 65536KB 64位IO格式: %I64d & %I64u

提交 状态

已开启划词翻译

问题描述

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. 

输入

* 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. 

输出

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

样例输入

3 31 22 12 3

样例输出

1

提示

Cow 3 is the only cow of high popularity. 

来源

USACO 2003 Fall

大意：输入n个牛m个有序对，（A，B）表示A认为B是红人。。。。最后求被所有其他牛都认为是红人的牛的个数。

思路：强连通分量的分解，先顺序dfs一次再逆序dfs一次，因为强连通分量2次都是能相互连通的，所以求出了强连通分量（既是都互相认为是红人的）。最后再拓扑排序最后的一个强连通分量里面牛个数就是最后的答案（复杂度O(m+n)）代码参考挑战程序设计竞赛

#include<iostream>#include<cstdio>#include<vector>#include<algorithm>#include<cstring>using namespace std;const int maxn=50005;int N,M,V;int A[maxn],B[maxn];vector<int> G[maxn];vector<int> rG[maxn];vector<int> vs;//后序遍历顺序bool used[maxn];int cmp[maxn];//拓扑排序void add(int f,int t){    G[f].push_back(t);    rG[t].push_back(f);}void dfs(int v){    used[v]=true;    for(int i=0;i<G[v].size();i++)    {        if(!used[G[v][i]]) dfs(G[v][i]);    }    vs.push_back(v);}void rdfs(int v,int k){    used[v]=true;    cmp[v]=k;    for(int i=0;i<rG[v].size();i++)    {        if(!used[rG[v][i]]) rdfs(rG[v][i],k);    }}int scc(){    memset(used,0,sizeof(used));    vs.clear();    for(int v=0;v<V;v++)    {        if(!used[v]) dfs(v);    }    memset(used,0,sizeof(used));    int k=0;    for(int i=vs.size()-1;i>=0;i--)    {        if(!used[vs[i]]) rdfs(vs[i],k++);    }    return k;}int main(){    cin>>N>>M;    V=N;    for(int i=0;i<M;i++)    {        cin>>A[i]>>B[i];        add(A[i]-1,B[i]-1);    }    int n=scc();    int u=0,num=0;    for(int v=0;v<V;v++)    {        if(cmp[v]==n-1)        {            u=v;            num++;        }    }    memset(used,0,sizeof(used));    rdfs(u,0);    for(int v=0;v<V;v++)    {        if(!used[v])        {            num=0;            break;        }    }    cout<<num<<endl;    return 0;}