POJ——T3259 Wormholes

http://poj.org/problem?id=3259

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 50692		Accepted: 18750

发现虫洞的的JOhn~开始他的神秘之~~~~

就是给你m条带权双向边，w条带权单向边

可以将单边制成负的，SFPA求一下是否有负环

 #include <cstring>
 #include <cstdio>

 #define MIN(a,b) ( a<b ?a :b)

 using namespace std;

 const int N();
 int t,n,m,c,u,v,w;
 int head[N],sumedge;
 struct Edge
 {
     int to,next,cost;
     Edge(int to=,int next=,int cost=):
         to(to),next(next),cost(cost){}
 }edge[N<<];

 void ins(int from,int to,int cost)
 {
     edge[++sumedge]=Edge(to,head[from],cost);
     head[from]=sumedge;
 }

 int if_YES,vis[N],dis[N];

 void SPFA(int now)
 {
     vis[now]=;
     if(if_YES) return ;
     for(int i=head[now];i;i=edge[i].next)
     {
         int go=edge[i].to;
         if(dis[go]>dis[now]+edge[i].cost)
         {
             if(vis[go])
             {
                 if_YES=;
                 break;
             }
             dis[go]=dis[now]+edge[i].cost;
             SPFA(go);
         }
     }
     vis[now]=;
 }

 void init()
 {
     if_YES=sumedge=;
     memset(vis,,sizeof(vis));
     memset(dis,,sizeof(dis));
     memset(head,,sizeof(head));
 }

 int main()
 {
     scanf("%d",&t);
     for(;t;t--)
     {    init();
         scanf("%d%d%d",&n,&m,&c);
         for(;m;m--)
         {
             scanf("%d%d%d",&u,&v,&w);
             ins(u,v,w); ins(v,u,w);
         }
         for(;c;c--)
         {
             scanf("%d%d%d",&u,&v,&w);
             ins(u,v,-w);
         }
         for(int i=;i<=n;i++)
         {
             SPFA(i);
             if(if_YES) break;
         }
         if(if_YES) printf("YES\n");
         else        printf("NO\n");
     }
     return ;
 }