Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 6628 Accepted: 2097

简单最短路问题，用bellman判断一下是否有负环的存在即可

代码：

 

 1 #include<stdio.h>
 2 #include<string.h>
 3  struct node
 4 {
 5     int a,b;
 6     double rab,cab;
 7 }g[205];
 8  int n,m;
 9 double dist[105];
10 int relax(int u,int v,double rab,double cab)
11 {
12     if(dist[v]<(dist[u]-cab)*rab)
13     {
14         dist[v]=(dist[u]-cab)*rab;
15         return 1;
16     }
17     return 0;
18 }
19 int bellman(int v0,double v)
20 {
21     int i,j;
22     for(i=1;i<=n;i++)
23         dist[i]=-1;
24     dist[v0]=v;
25     int flag;
26     for(i=1;i<n;i++)
27     {
28         flag=0;
29         for(j=1;j<=2*m;j++)
30         {
31             if(relax(g[j].a,g[j].b,g[j].rab,g[j].cab))
32             {
33                 flag=1;
34             }
35         }
36         if(!flag)
37             break;
38     }
39     for(j=1;j<=m*2;j++)
40     {
41         if(relax(g[j].a,g[j].b,g[j].rab,g[j].cab))
42             return 1;
43     }
44     return 0;
45 }
46 int main()
47 {
48     double v,rab,cab,rba,cba;
49     int i,s,a,b;
50     scanf("%d%d%d%lf",&n,&m,&s,&v);
51     for(i=1;i<=m;i++)
52     {
53         scanf("%d%d%lf%lf%lf%lf",&a,&b,&rab,&cab,&rba,&cba);
54         g[i].a=a;g[i].b=b;
55         g[i].rab=rab;g[i].cab=cab;
56         g[i+m].a=b;g[i+m].b=a;
57         g[i+m].rab=rba;g[i+m].cab=cba;
58     }
59     if(bellman(s,v))
60         printf("YES\n");
61     else
62         printf("NO\n");
63     return 0;
64 }