poj 3259 Wormholes

这题和poj 1860很像，不过那个是求上升回路，这题是求下降回路。
 1 #include <stdio.h>
 2 #include <string.h>
 3 int u[5400],v[5400],w[5400];
 4 int d[505];
 5 int n,m1,m2;
 6 bool Bellman()
 7 {
 8     int i,j;
 9     memset(d,0x3f,sizeof(d));
10     d[1] = 0;
11     for(j = 1; j < n; j++)
12     {
13         bool ok = 1;
14         for(i = 0; i < 2*m1+m2; i++)
15         if(d[u[i]] + w[i] < d[v[i]])
16         {
17             d[v[i]] = d[u[i]] + w[i];
18             ok = 0;
19         }
20         if(ok) break;
21     }
22     for(i = 0; i < 2*m1+m2; i++)
23         if(d[u[i]] + w[i] < d[v[i]])
24             return 1;
25     return 0;
26 }
27 int main()
28 {
29     int F,i;
30     scanf("%d",&F);
31     while(F--)
32     {
33         scanf("%d%d%d",&n,&m1,&m2);
34         for(i = 0; i < 2*m1; i += 2)
35         {
36             scanf("%d%d%d",&u[i],&v[i],&w[i]);
37             u[i+1] = v[i];
38             v[i+1] = u[i];
39             w[i+1] = w[i];
40         }
41         for(i = 2*m1; i < 2*m1+m2; i++)
42         {
43             scanf("%d%d%d",&u[i],&v[i],&w[i]);
44             w[i] = -w[i];
45         }
46         if(Bellman())
47             printf("YES\n");
48         else printf("NO\n");
49     }
50     return 0;
51 }