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Re:一次A过,留下代码做纪念~(bellman-ford)

Posted by 20171044 at 2018-08-13 15:51:39 on Problem 3259
In Reply To:一次A过,留下代码做纪念~(bellman-ford) Posted by:sunyi740 at 2016-03-24 22:17:29
> #include<iostream>
> #include<cstring>
> using namespace std;
> int dis[6000];
> int mix=10005;
> int n,m,w;
> struct edge
> {
>     int u;
>     int v;
>     int t;
> };
> edge a[6000];
> 
> void bellman()
> {
>     memset(dis,mix,sizeof(dis));
>     dis[1]=0;
>     for(int i=1;i<n-1;i++)
>         for(int j=1;j<=2*m+w;j++)
>         {
>             if(dis[a[j].u]+a[j].t<dis[a[j].v])
>                 dis[a[j].v]=dis[a[j].u]+a[j].t;
>         }
>     int flag=0;
>     for(int j=1;j<=2*m+w;j++)
>     {
>         if(dis[a[j].u]+a[j].t<dis[a[j].v])
>         {
>             flag=1;
>             break;
>         }
>     }
>     if(flag==1) cout<<"YES"<<endl;
>     else cout<<"NO"<<endl;
>     return;
> }
> 
> int main()
> {
>     int f;
>     cin>>f;
>     while(f--)
>     {
>         cin>>n>>m>>w;
>         int i;
>         for(i=1;i<=2*m-1;i=i+2)
>         {
>             cin>>a[i].u>>a[i].v>>a[i].t;
>             a[i+1].u=a[i].v;
>             a[i+1].v=a[i].u;
>             a[i+1].t=a[i].t;
>         }
>         for(i=2*m+1;i<=2*m+w;i++)
>         {
>             cin>>a[i].u>>a[i].v>>a[i].t;
>             a[i].t=-a[i].t;
>         }
>         bellman();
>     }
>     return 0;
> }

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