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！

 1 #include <iostream>

 2 #include <stdio.h>

 3 #include <string.h>

 4 #include <cstring>

 5 #include<map>

 6 using namespace std;

 7 

 8 map<string,int>name;

 9 const int INF = 1000000;

10 const int MAXSIZE = 1005;

11 const int MAXN = 30;

12 double g[MAXN][MAXN];

13 

14 //dis[][]记录任意两点间的最短路径，初始的dis[][]记录直接路径

15 void floyed(double dis[][MAXN],int n){//节点从1~n编号

16     int i,j,k;

17     for(k = 1; k <= n; k++)

18        for(i = 1; i <= n; i++)

19          for(j = 1; j <= n; j++)

20              if(dis[i][j] < dis[i][k] * dis[k][j])

21                  dis[i][j] = dis[i][k] * dis[k][j];

22 }

23 int main(){

24     int n,i,m,j;

25     string str1,str2;

26     double r;

27     int iCase = 0;

28     while(scanf("%d",&n),n){

29         iCase++;

30         for(i = 1; i <= n; i++){

31             cin>>str1;

32             name[str1] = i;

33         }

34         for(i = 1; i <= n; i++)

35            for(j = 1; j <= n; j++){

36                if(i == j) g[i][j] = 1;

37                else g[i][j] = 0;

38            }

39         scanf("%d",&m);

40         while(m--){

41             cin>>str1>>r>>str2;

42             g[name[str1]][name[str2]] = r;

43         }

44         

45         floyed(g,n);

46         

47         bool flag = false;

48         for(i = 1; i <= n; i++){

49           if(g[i][i] > 1){

50             flag = true;

51             break;

52           }

53         }

54         if(flag)  printf("Case %d: Yes\n",iCase);

55         else printf("Case %d: No\n",iCase);

56     }

57     return 0;

58 }