Floyd算法

Floyd算法是穷举图中所有中转点，不断优化起点到终点的距离，可以用来求多源最短路问题和传递闭包。

H——Cow Contest

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 determine

#include<iostream>
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<cstdio>
#include<vector>
using namespace std;
#define MAXN 101
#define INF 0x3f3f3f3f
/*
9 03
9 11
学习Floyd算法
*/
bool win[MAXN][MAXN];
int n,m;
int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        int x,y;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);
            win[x][y] = true;
        }
        for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++)
                    if(win[i][k]&&win[k][j])
                        win[i][j] = true;
        int ans = 0,t;
        for(int i=1;i<=n;i++)
        {
            for(t=1;t<=n;t++)
            {
                if(t==i) continue;
                if(!win[i][t]&&!win[t][i])
                    break;
            }
            if(t>n) ans++;
        }
        cout<<ans<<endl;
    }
    return 0;
}

I - Arbitrage

 相当于判断有无正环，Floyd.

#include<iostream>
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<map>
using namespace std;
#define MAXN 31
#define INF 0x3f3f3f3f
/*
map<string,int> 确定位置 超时
12 23
12 47
*/
int n,m;
map<string,int> Mp;
double g[MAXN][MAXN];
void Floyd()
{
    for(int k=0;k<n;k++)
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                g[i][j] = max(g[i][j],g[i][k]*g[k][j]);
            }
}
int main()
{
    int cnt = 0;
    while(scanf("%d",&n),n)
    {
        Mp.clear();
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                if(j==i) g[i][j] = 1.0;
                else g[i][j] = 0.0;
            }
        string str,t1,t2;
        double rate;
        for(int i=0;i<n;i++)
        {
            cin>>str;
            Mp[str] = i;
        }
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            cin>>t1>>rate>>t2;
            g[Mp[t1]][Mp[t2]] = rate;
        }
        Floyd();
        bool f= false;
        printf("Case %d: ",++cnt);
        for(int i=0;i<n;i++)
            if(g[i][i]>1.0)
            {
                f = true;
                cout<<"Yes
";
                break;
            }
        if(!f)    
            cout<<"No
";
    }
    return 0;
}