POJ 3660 Cow Contest

POJ 3660 Cow Contest

 

Cow Contest

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:7628		Accepted:4243

 

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 conducting CTED in several head-to-head rounds, each between two cows. If cowAHas a greater skill level than cowB(1 ≤A≤N; 1 ≤B≤N;A=B), Then cowAWill always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the resultsM(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 contradic.pdf.

Input

* Line 1: Two space-separated integers:NAndM
* 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:AAndB

Output

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

Sample Input

5 54 34 23 21 22 5

Sample Output

2

// The order in which ox A can defeat ox B is determined.

// Record the degree of each ox, if equal to the N-1 indicates the order can be determined.

 

 

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     
      #include #define N 109using namespace std;int mp[N][N];int n,m;int a,b;int deg[N];int main(){    while(~scanf("%d %d",&n,&m))    {        memset(deg,0,sizeof deg);        memset(mp,0,sizeof mp);        for(int i=1;i<=m;i++)        {            scanf("%d %d",&a,&b);            mp[a][b]=1;        }        for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)        for(int k=1;k<=n;k++)        if(mp[j][i] && mp[i][k])        mp[j][k]=1;        for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)        {            if(mp[i][j])            {                deg[i]++;                deg[j]++;            }        }        int ans=0;        for(int i=1;i<=n;i++)        if(deg[i]==n-1)        ans++;        printf("%d\n",ans);    }    return 0;}