poj2240 Bellman-ford最短路求最大回路

Description

Input

Output

Sample Input

3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0

Sample Output

Case 1: Yes
Case 2: No

Source   本题是让求是否可以套汇的.(在货币的兑换过程是否可以获利)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define INF 0x3f3f3f3f

using namespace std;

struct node
{
    double cij;
    int ci,cj;
} edge[10000];
int n,m;
char s[40][100];
char s1[100],s2[100];
double Max[10000];    //double 类型
int path[10000];
int flag=0;
int Bellman(int u0)
{
	flag=0;
    memset(Max,0,sizeof(Max));   //初始化为最小
    Max[u0]=1;
    for(int i=1; i<=n; i++)  //构成环所以要递推n次
    {
        for(int j=0; j<m; j++)
        {
            if(Max[edge[j].ci]*edge[j].cij>Max[edge[j].cj])
                Max[edge[j].cj]=Max[edge[j].ci]*edge[j].cij;
        }
    }
    if(Max[u0]>1.0)
        flag=1;
}
int main()
{
    int Case=1;
    while(~scanf("%d",&n))
    {

        if(n==0)
            break;
        for(int i=0; i<n; i++)
        {
            scanf("%s",s[i]);
        }

        scanf("%d",&m);
        int j,k;
        for(int i=0; i<m; i++)
        {
            double d;
            scanf("%s %lf %s",s1,&d,s2);
            for(j=0; strcmp(s1,s[j]); j++)
                ;
            for(k=0; strcmp(s2,s[k]); k++)
                ;
            edge[i].ci=j;
            edge[i].cj=k;
            edge[i].cij=d;
        }
        for(int i=0; i<n; i++)
        {
            Bellman(i);
            if(flag)
                break;
        }
        if(flag)
            printf("Case %d: Yes\n",Case++);
        else
            printf("Case %d: No\n",Case++);
    }
}