HDOJ 1217 Floyed Template

解题思路：
1.map简单应用
2.Floyd算法的变形，之后判断dis[i][i]，如果大于1，则存在利润！

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cstring>
#include<map>
using namespace std;

map<string,int>name;
const int INF = 1000000;
const int MAXSIZE = 1005;
const int MAXN = 30;
double g[MAXN][MAXN];

//dis[][]记录任意两点间的最短路径，初始的dis[][]记录直接路径
void floyed(double dis[][MAXN],int n){//节点从1~n编号
    int i,j,k;
    for(k = 1; k <= n; k++)
       for(i = 1; i <= n; i++)
         for(j = 1; j <= n; j++)
             if(dis[i][j] < dis[i][k] * dis[k][j])
                 dis[i][j] = dis[i][k] * dis[k][j];
}
int main(){
    int n,i,m,j;
    string str1,str2;
    double r;
    int iCase = 0;
    while(scanf("%d",&n),n){
        iCase++;
        for(i = 1; i <= n; i++){
            cin>>str1;
            name[str1] = i;
        }
        for(i = 1; i <= n; i++)
           for(j = 1; j <= n; j++){
               if(i == j) g[i][j] = 1;
               else g[i][j] = 0;
           }
        scanf("%d",&m);
        while(m--){
            cin>>str1>>r>>str2;
            g[name[str1]][name[str2]] = r;
        }
        
        floyed(g,n);
        
        bool flag = false;
        for(i = 1; i <= n; i++){
          if(g[i][i] > 1){
            flag = true;
            break;
          }
        }
        if(flag)  printf("Case %d: Yes
",iCase);
        else printf("Case %d: No
",iCase);
    }
    return 0;
}