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(2521);
int t,n,m,c,u,v,w;
int head[N],sumedge;
struct Edge
{
    int to,next,cost;
    Edge(int to=0,int next=0,int cost=0):
        to(to),next(next),cost(cost){}
}edge[N<<1];

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]=1;
    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=1;
                break;
            }
            dis[go]=dis[now]+edge[i].cost;
            SPFA(go);
        }
    }
    vis[now]=0;
}

void init()
{
    if_YES=sumedge=0;
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    memset(head,0,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=1;i<=n;i++)
        {
            SPFA(i);
            if(if_YES) break;
        }
        if(if_YES) printf("YES
");
        else        printf("NO
");
    }
    return 0;
}