POJ 2240 Arbitrage

解题思路: Bellman-Ford算法，求解负权回路问题，考虑到图的不连通性，添加一个超级初始点，到各顶点汇率均为1：

#include <iostream>
#include <map>
#include <string>
using namespace std;

int main()
{
    int N, M , L = 1;
    double rate;
    string curStr, curStr1, curStr2;
    map<string, int>name;
    double dist[305];

    struct  
    {
        int u,v;
        double rate;
    }edge[1000];

    while (cin >> N && N)
    {
        fill(&dist[0], &dist[304], 1000.0);
        for (int i = 1; i <= N; i++)
        {
            cin >> curStr;
            name[curStr] = i;
        }
        cin >> M;
        for (int i = 0; i < M; i++)
        {
            cin >> curStr1 >> rate >> curStr2;
            edge[i].u = name[curStr1];
            edge[i].v = name[curStr2];
            edge[i].rate = rate;
        }
        bool isChanged = true;
        for (int i = 0; i < N && isChanged; i++)
        {
            isChanged = false;
            for (int j = 0; j < M; j++)
            {
                if (dist[edge[j].v] < dist[edge[j].u] * edge[j].rate)
                {
                    dist[edge[j].v] = dist[edge[j].u] * edge[j].rate;
                    isChanged = true;
                }
            }
        }
        name.clear();
        cout <<"Case " << L << (isChanged ? (": Yes") : (": No")) << endl;
        L++;
    }
    return 0;
}