DFS 确定排名

Cow Contest

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2

思路：

              如果要知道一个点的排名是否确定，我们可以这样考虑，找到排名比他高的和比他低的（不一定要知道比它高多少名或者低多少名，只要能确定大小即可）；

            先把给的每个点连成一个有向图（我这里的方向是大的指向向的），之后我们判断每一个点向上和向下可以找到多少个点，如果总的找到了n-1个点，说明这个点的排名就是可以知道的，记录下来。再继续寻找下一个点；

这里是用DFS搜索的；

#include<vector>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
#define INF 0x3f3f3f
int n,m,mp[110][110],flag,vis[110];
struct point{
    vector<int> into;            //该点的入度
    vector<int> out;               //该点的出度
}pt[110];

void DFS_out(int k){
    if(vis[k])return;
    vis[k]=1;
    for(int i=0;i<pt[k].out.size();i++){
        if(vis[pt[k].out[i]])continue;
        flag++;
        DFS_out(pt[k].out[i]);
    }
}

void DFS_into(int k){
    if(vis[k])return;
    vis[k]=1;
    for(int i=0;i<pt[k].into.size();i++){
        if(vis[pt[k].into[i]])continue;
        flag++;
        DFS_into(pt[k].into[i]);
    }
}
int main(){
    scanf("%d%d",&n,&m);
    int a,b;

    for(int i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
            pt[a].out.push_back(b);
            pt[b].into.push_back(a);
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        flag=0;
        memset(vis,0,sizeof(vis));            //记得要初始化！！！
        DFS_out(i);
        memset(vis,0,sizeof(vis));
        DFS_into(i);
        if(flag==n-1)ans++;
    }

    printf("%d\n",ans);
    return 0;
}