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1 #include <stdio.h> 2 #include <string.h> 3 int n,m,ans; 4 char str[40][50]; 5 double dis[40]; 6 struct f 7 { 8 int x,y; 9 double z; 10 }rate[1000]; 11 int num(char s[]) //输出对应货币的编号 12 { 13 int i; 14 for (i = 1; i <= n; i ++) 15 if (strcmp(str[i],s)==0) 16 break; 17 return i; 18 } 19 void add(int a,int b,double c) 20 { 21 rate[ans].x = a; 22 rate[ans].y = b; 23 rate[ans].z = c; 24 ans ++; 25 } 26 bool Bellman_ford() 27 { 28 int i,j; 29 for (i = 1; i <= n; i ++) 30 dis[i] = 0; 31 dis[1] = 100; 32 for (i = 1; i < n; i ++) 33 for (j = 0; j < ans; j ++) 34 if (dis[rate[j].y] < dis[rate[j].x]*rate[j].z) 35 dis[rate[j].y] = dis[rate[j].x]*rate[j].z; 36 for (j = 0; j < ans; j ++) 37 if (dis[rate[j].y] < dis[rate[j].x]*rate[j].z) 38 return true; 39 return false; 40 } 41 int main () 42 { 43 int i,j,f = 1; 44 double c; 45 char ch1[100],ch2[100]; 46 while (~scanf("%d",&n)) 47 { 48 ans = 0; 49 if (n == 0) 50 break; 51 for (i = 1; i <= n; i ++) 52 scanf(" %s",str[i]); 53 54 55 scanf("%d",&m); 56 for (j = 0; j < m; j ++) 57 { 58 scanf("%s %lf %s",ch1,&c,ch2); 59 int a = num(ch1); 60 int b = num(ch2); 61 add(a,b,c); 62 } 63 64 int flag = 0; 65 printf("Case %d: ",f ++); 66 if (Bellman_ford()) 67 printf("Yes\n"); 68 else 69 printf("No\n"); 70 } 71 return 0; 72 }
Floyd算法：
1 //Memory Time 2 //276K 79MS 3 4 #include <iostream> 5 #include<map> 6 #include<string> 7 using namespace std; 8 9 const int inf=10000; //无限大 10 int n; //货币种类 11 int m; //兑换方式 12 13 map<string,int>STL; //建立一个 使字符串与整数有一一对应关系 的容器STL,以便利用邻接矩阵存储数据 14 15 double rate; 16 char str[50],str1[50],str2[50]; 17 double dist[31][31]; 18 19 int i,j,k; 20 21 void floyd(void) 22 { 23 for(k=1;k<=n;k++) 24 for(i=1;i<=n;i++) 25 for(j=1;j<=n;j++) 26 if(dist[i][j] < dist[i][k] * dist[k][j]) //变形的最大路径，变"+"为"*" 27 dist[i][j] = dist[i][k] * dist[k][j]; 28 return; 29 } 30 31 int main(void) 32 { 33 int cases=1; 34 while(cases) 35 { 36 memset(dist,inf,sizeof(dist)); 37 38 39 cin>>n; 40 if(!n)break; 41 42 for(i=1;i<=n;i++) 43 { 44 cin>>str; 45 STL[str]=i; //将输入的货币从1到n依次编号 46 dist[i][i]=1; //到自身的转换率默认为1，但通过floyd可能会被改变 47 //有向图的顶点（一般）存在环 48 } 49 50 cin>>m; 51 for(i=1;i<=m;i++) 52 { 53 cin>>str1>>rate>>str2; 54 dist[STL[str1]][STL[str2]]=rate; //构造图 55 } 56 57 floyd(); 58 59 int flag=false; 60 for(i=1;i<=n;i++) 61 if(dist[i][i]>1) 62 { 63 flag=true; 64 break; 65 } 66 if(flag) 67 cout<<"Case "<<cases++<<": Yes"<<endl; 68 else 69 cout<<"Case "<<cases++<<": No"<<endl; 70 } 71 return 0; 72 }
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原文地址：http://www.cnblogs.com/yoke/p/5875857.html
