POJ 3660 Cow Contest (Floyd transfer closure)

POJ 3660 Cow Contest (Floyd transfer closure)

Cow Contest

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:7443		Accepted:4133

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

Source

USACO 2008 January Silver

Question link: http://poj.org/problem? Id = 3660

N people, m groups, and relationship a B indicate that a ranks before B, and the number of people who are asked to rank has been determined.

Question Analysis: use Floyd to calculate the transfer closure. a is before B and B is before c, then a is before c. If the relationship between the position of each Floyd vertex and other vertices has been determined, the ranking is determined.

#include 
 
  #include 
  
   int const MAX = 105;bool ok[MAX][MAX];int n, m;void Floyd(){    for(int k = 1; k <= n; k++)        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)                if(ok[i][k] && ok[k][j])                    ok[i][j] = true;}int main(){    scanf("%d %d", &n, &m);    memset(ok, false, sizeof(ok));    for(int i = 0; i < m; i++)    {        int a, b;        scanf("%d %d", &a, &b);        ok[a][b] = true;    }    Floyd();    int ans = 0;    for(int i = 1; i <= n; i++)    {        bool flag = true;        for(int j = 1; j <= n; j++)        {            if(i == j)                continue;            if(!ok[i][j] && !ok[j][i])            {                flag = false;                break;            }        }        if(flag)            ans ++;    }    printf("%d\n", ans);}