poj 2240 Arbitrage

又是套汇为题，初始化d数组为1就可以了，用bellman—ford算法。

#include<stdio.h>
#include<string.h>

int n, m, num, u[2000], v[2000];
double w[2000], d[50], ra;
char s[50][100], s1[100], s2[100];

void solve()
{
    for(int i = 0; i <= n; i ++) d[i] = 1;
    for(int k = 0; k < n-1; k ++)
        for(int i = 0; i < m; i ++)
        {
            int x = u[i], y = v[i];
            if(d[y] < d[x]*w[i]) d[y] = d[x]*w[i];
        }    
    int flag = 0;
    for(int i = 0; i < m; i++)
    {
        int x = u[i],y = v[i];
        if(d[y] < d[x]*w[i]) flag = 1;
    }
    if(flag == 1) printf("Case %d: Yes\n",num++);
    else printf("Case %d: No\n",num++);
}

void init()
{
    while(scanf("%d",&n))
    {
        if(n == 0) break;
        for(int i = 0; i < n; i ++)
            scanf("%s",s[i]);
        scanf("%d",&m);
        for(int i = 0; i < m; i ++)
        {
            scanf("%s",s1);
            scanf("%lf",&ra);
            scanf("%s",s2);
            int a, b;
            for(int j = 0; j < n; j ++)
            {
                if(!strcmp(s[j], s1)) a = j;
                if(!strcmp(s[j], s2)) b = j;
            }
            u[i] = a;
            v[i] = b;
            w[i] = ra;
        }
        solve();
    }    
}

int main()
{
    num = 1;
    init();
    return 0;
}