UVA 436Arbitrage (II)【floyd】

来源：互联网发布：华夏风云武将数据编辑：程序博客网时间：2024/06/05 04:27

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=377

题意：硬币兑换，能不能兑换一圈使得手中的钱变多。
floyd传递闭包。

代码：

#include <stdio.h>#include <iostream>#include <algorithm>#include <string.h>#include <queue>#include <math.h>#include <map>#include <string>using namespace std;int n, m;map<string, int> MP;double a[50][50];int main(){    string s,y;    double tmp;    int cases = 1;    while (cin >> n && n)    {        memset(a, 0, sizeof(a));        MP.clear();        for (int i = 1; i <= n; i++)        {            cin >> s;            MP[s] = i;        }        for (int i = 1; i <= n; i++)            a[i][i] = 1.0;        cin >> m;        for (int i = 1; i <= m; i++)        {            cin >> s >> tmp >> y;            a[MP[s]][MP[y]] = tmp;        }        for (int k = 1; k <= n; k++)            for (int i = 1; i <= n; i++)                for (int j = 1; j <= n; j++)                {                    if (a[i][j] < a[i][k] * a[k][j])                        a[i][j] = a[i][k] * a[k][j];                }        int ok = 0;        for (int i = 1; i <= n; i++)            if (a[i][i] > 1.0)            {                ok = 1;                break;            }        printf("Case %d: ",cases++);        if (ok) printf("Yes\n");        else printf("No\n");    }    return 0;}

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