POJ 2240 Arbitrage

题目大意:跟我上次做的那道感觉一模一样，上次链接:https://www.cnblogs.com/ducklu/p/9231563.html

解题思路:Bellman_Ford判断有无正环（对了，第一次RE了，数组要开大点，我开始只开了40）

#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
#define INF 1000000
#define MAXVERTEXNUM 1000
using namespace std;

int Nv, Ne, num;
struct Edge
{
    int a, b;
    double Weight;
}E[MAXVERTEXNUM];
double dist[MAXVERTEXNUM];

bool Bellman_ford()
{
    for (int i = 1; i <= Nv; ++i)
    {
        bool flag = false;
        for (int j = 0; j < num; ++j)
        {
            if (dist[E[j].b] < dist[E[j].a] * E[j].Weight)
            {
                dist[E[j].b] = dist[E[j].a] * E[j].Weight;
                flag = true;
            }
        }
        if (!flag)
            break;
    }

    for (int j = 0; j < num; ++j)
        if (dist[E[j].b] < dist[E[j].a] * E[j].Weight)
            return true;

    return false;
}

int main()
{
//    freopen("test.txt", "r", stdin);
    int cas = 0;
    while (true)
    {
        num = 0;
        map<string, int> m;
        memset(dist, 0, sizeof(dist));
        dist[1] = 1;

        cin >> Nv;
        if (!Nv)
            return 0;
        cas++;

        for (int i = 1; i <= Nv; ++i)
        {
            string temp;
            cin >> temp;
            m.insert( {temp, i} );
        }

        cin >> Ne;

        for (int i = 1; i <= Ne; ++i)
        {
            string s1, s2;
            double W;
            cin >> s1 >> W >> s2;

            E[num].a = m[s1], E[num].b = m[s2];
            E[num++].Weight = W;
        }

        if (Bellman_ford())
            cout << "Case " << cas << ": Yes" << endl;
        else
            cout << "Case " << cas << ": No" << endl;

//        getchar();
    }

    return 0;
}