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

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define INF 1000000
 5 #define MAXVERTEXNUM 505
 6 #define MAXEDGENUM 10000
 7 using namespace std;
 8 
 9 int Nv, Ne1, Ne2, num;
10 int dist[MAXVERTEXNUM];
11 struct E
12 {
13     int a, b, Weight;
14 }Edge[MAXEDGENUM];
15 
16 bool BellManFord()
17 {
18     for (int i = 1; i < Nv; ++i)
19     {
20         bool flag = false;
21         for (int j = 0; j < num; ++j)
22         {
23             if (dist[Edge[j].b] > dist[Edge[j].a] + Edge[j].Weight)
24             {
25                 dist[Edge[j].b] = dist[Edge[j].a] + Edge[j].Weight;
26                 flag = true;
27             }
28         }
29         if (!flag)
30             break;
31     }
32 
33     for (int i = 0; i < num; ++i)
34         if (dist[Edge[i].b] > dist[Edge[i].a] + Edge[i].Weight)
35             return true;
36 
37     return false;
38 }
39 
40 int main()
41 {
42     ios::sync_with_stdio(false);
43 
44     int cas;
45     cin >> cas;
46     while (cas--)
47     {
48         num = 0;
49         cin >> Nv >> Ne1 >> Ne2;
50         for (int i = 0; i < Ne1; ++i)
51         {
52             int V1, V2, W;
53             cin >> V1 >> V2 >> W;
54             Edge[num].a = V1, Edge[num].b = V2;
55             Edge[num++].Weight = W;
56             Edge[num].a = V2, Edge[num].b = V1;
57             Edge[num++].Weight = W;
58         }
59 
60         for (int i = 0; i < Ne2; ++i)
61         {
62             int V1, V2, W;
63             cin >> V1 >> V2 >> W;
64             Edge[num].a = V1, Edge[num].b = V2;
65             Edge[num++].Weight = -W;
66         }
67 
68         //init
69         for (int i = 1; i <= Nv; ++i)
70             dist[i] = INF;
71         dist[1] = 0;
72 
73         if (BellManFord())
74             cout << "YES" << endl;
75         else
76             cout << "NO" << endl;
77     }
78 
79     return 0;
80 }