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 = ;
 const int MAXSIZE = ;
 const int MAXN = ;
 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 = ; k <= n; k++)
        for(i = ; i <= n; i++)
          for(j = ; 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 = ;
     while(scanf("%d",&n),n){
         iCase++;
         for(i = ; i <= n; i++){
             cin>>str1;
             name[str1] = i;
         }
         for(i = ; i <= n; i++)
            for(j = ; j <= n; j++){
                if(i == j) g[i][j] = ;
                else g[i][j] = ;
            }
         scanf("%d",&m);
         while(m--){
             cin>>str1>>r>>str2;
             g[name[str1]][name[str2]] = r;
         }

         floyed(g,n);

         bool flag = false;
         for(i = ; i <= n; i++){
           if(g[i][i] > ){
             flag = true;
             break;
           }
         }
         if(flag)  printf("Case %d: Yes\n",iCase);
         else printf("Case %d: No\n",iCase);
     }
     return ;
 }