POJ 3259 Wormholes

题意:

农夫 FJ 有 N 块田地【编号 1...n】 (1<=N<=500)
田地间有 M 条路径 【双向】(1<= M <= 2500)
同时有 W 个孔洞,可以回到以前的一个时间点【单向】(1<= W <=200)
问：FJ 是否能在田地中遇到以前的自己

解题思路:跟我昨天做的一个题思路差不多，这道题判断有无负权环就行了，昨天做的链接https://www.cnblogs.com/ducklu/p/9231563.html

#include <iostream>
#include <cstring>
#include <algorithm>
#define INF 1000000
#define MAXVERTEXNUM 505
#define MAXEDGENUM 10000
using namespace std;

int Nv, Ne1, Ne2, num;
int dist[MAXVERTEXNUM];
struct E
{
    int a, b, Weight;
}Edge[MAXEDGENUM];

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

    for (int i = 0; i < num; ++i)
        if (dist[Edge[i].b] > dist[Edge[i].a] + Edge[i].Weight)
            return true;

    return false;
}

int main()
{
    ios::sync_with_stdio(false);

    int cas;
    cin >> cas;
    while (cas--)
    {
        num = 0;
        cin >> Nv >> Ne1 >> Ne2;
        for (int i = 0; i < Ne1; ++i)
        {
            int V1, V2, W;
            cin >> V1 >> V2 >> W;
            Edge[num].a = V1, Edge[num].b = V2;
            Edge[num++].Weight = W;
            Edge[num].a = V2, Edge[num].b = V1;
            Edge[num++].Weight = W;
        }

        for (int i = 0; i < Ne2; ++i)
        {
            int V1, V2, W;
            cin >> V1 >> V2 >> W;
            Edge[num].a = V1, Edge[num].b = V2;
            Edge[num++].Weight = -W;
        }

        //init
        for (int i = 1; i <= Nv; ++i)
            dist[i] = INF;
        dist[1] = 0;

        if (BellManFord())
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }

    return 0;
}