poj3660

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 (1≤A≤N) has a greater skill level than cow B (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≤4500) 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.

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

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

题意:

n (1≤n≤100)个人参加比赛, 有m (1≤m≤4500)条实力信息, 对于每一条信息A B(1≤A,B≤n, A!=B)表示A的实力比B强.

在保证输入信息不矛盾的情况下, 求最后有多少个人可以确定自己的排名.

思路:

Floyd的模板题.

如果$v_i$与$v_k$有关系且$v_k$和$v_j$有关系并且三者的关系相同那么就进行松弛.

最后再计算有多少个人和其他n-1个人都有关系.

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#include <cstring>
using namespace std;
const int maxn = 1e3 + 5;

int n, m;
int dis[maxn][maxn]; 

int ()
{
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                if(dis[i][k] != 0 && dis[i][k] == dis[k][j])
                    dis[i][j] = dis[i][k];

    int ans = 0;
    for(int i = 1; i <= n; i++)
    {
        int tmp = 0;
        for(int j = 1; j <= n; j++)
            if(i == j)
                continue;
            else if(dis[i][j] != 0)
                tmp++;
            else
                break;
        if(tmp == n - 1)
            ans++;
    }
    return ans;
}

int main()
{
    ios_base::sync_with_stdio(0);cin.tie();
    cin >> n >> m;
    memset(dis, 0, sizeof(dis));
    for(int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        dis[a][b] = -1;
        dis[b][a] = 1;
    }
    cout << floyd() << endl;
    return 0;
}