Poj 2168 popular cows

Description

Every cow's dream is to becomethe 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 (, b) that tell you that cow a thinks that cow B is popular. since popularity
Istransitive, if a thinks B is popular and B thinks C is popular, then a willalso think that C is
Popular, even if this is not explicitly specified by an Ordered Pair in theinput. Your task is to compute the number of cows that are considered popularby every other cow.

Input

* Line 1: two space-separatedintegers, N and m

* Lines 2 .. 1 + M: two space-separated numbers a and B, meaning that a thinks B ispopular.

Output

* Line 1: A single integerthat 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

Topic Introduction: (a, B) indicates that a thinks B is popular, and each ox thinks that he is popular. If (a, B) AND (B, c), then (a, c ). That is, a considers B to be popular, and B thinks C to be popular. Then a also thinks C is popular. The problem is, find out how many cows are welcomed by all the cows.

Method: kosaraju algorithm. Actually, I don't know much about it ..... It was evening to solve this problem. This algorithm is learned only when this question is done. Look at this algorithm to see more than two points ..... I understand the algorithm, but I still don't know the principle ..... What is to be learned?

 

#include<iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;vector<int> v1[10010], v2[10010];vector<int> s;int num[10010], sum[10010], mark[10010];int counter;void DFS1(int x){    mark[x] = 1;for (int i = 0; i < v1[x].size(); ++i)        {if (!mark[v1[x][i]])                {DFS1(v1[x][i]);                }}s.push_back(x);};void DFS2(int x){    num[x] = counter;  sum[counter]++; mark[x] = 1;for (int i = 0; i < v2[x].size(); ++i)           {if (!mark[v2[x][i]])                   {DFS2(v2[x][i]);}}};void SUM(int n){    memset(mark, 0, sizeof(mark)); for (int i = 1; i <= n; i++)    {for (int j = 0; j < v1[i].size();j++)        {int x = v1[i][j];            if (num[i] != num[x]){mark[num[i]] = 1;}}    }int flag = 0, ans;for (int i = 1; i <= counter; i++){     if (!mark[i])       {ans = i;           flag++;       }}if (flag == 1){printf("%d\n", sum[ans]);}else{printf("0\n");}};int main(){int n, m, a, b, i, j;scanf("%d%d",&n,&m);for (i = 0;i<m;i++){scanf("%d%d", &a, &b);v1[a].push_back(b);v2[b].push_back(a);}for (i = 1;i<=n;i++){if (!mark[i]){DFS1(i);}}memset(mark, 0, sizeof(mark));counter = 0;for (i = s.size() - 1;i>=0;i--){if (!mark[s[i]]){counter++;DFS2(s[i]);}}SUM(n);return 0;}