H - Wormholes

Description
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ’s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input
Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.
Output
Lines 1.. F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).
Sample Input
2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8
Sample Output
NO
YES
Hint
For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

题意：给出普通的点，边和权值，以及虫洞的点，边，权值，问你是否能从某一点出发最后回到出发点，并且回到去的时间要在出发前。

先建边，普通边正值，虫洞边负值，接着对每一个点跑一次最短路(spfa)，如果存在负环，就是成立的情况。

代码：

#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <map>#include <iostream>#include <vector>#include <cmath>#include <cstdlib>using namespace std;const int MAXN=505;const int MAXM=800005;bool vis[MAXN];int cnt[MAXN];int dir[MAXN];struct Edge{    int v;    double w;};vector<Edge> edge[MAXN];void add(int u,int v,int w){    Edge tmp;    tmp.v=v;    tmp.w=w;    edge[u].push_back(tmp);}bool spfa(int from,int n){    memset(vis,false,sizeof(vis));    memset(dir,0x3f,sizeof(dir));    memset(cnt,0,sizeof(cnt));    dir[from]=0;    vis[from]=true;    queue<int>que;    while(!que.empty())    {        que.pop();    }    que.push(from);    while(!que.empty())    {        int now=que.front();        que.pop();        vis[now]=false;        for(int i=0;i<edge[now].size();i++)        {            if(dir[edge[now][i].v]>edge[now][i].w+dir[now])                {                    dir[edge[now][i].v]=edge[now][i].w+dir[now];                    if(!vis[edge[now][i].v])                    {                        que.push(edge[now][i].v);                        vis[edge[now][i].v]=true;                        if(++cnt[edge[now][i].v]>n)                            return false;                    }                }        }    }    return true;}int main (void){    int num;    cin>>num;    while(num--)    {        int N,M,W;        int s,e,t;        for(int i=1;i<=N;i++)        {            edge[i].clear();        }        scanf("%d %d %d",&N,&M,&W);        for(int i=1;i<=M;i++)        {            scanf("%d %d %d",&s,&e,&t);            add(s,e,t);            add(e,s,t);        }        for(int i=0;i<W;i++)        {            scanf("%d %d %d",&s,&e,&t);            add(s,e,-t);        }        bool re;        for(int i=1;i<=N;i++)        {            re=spfa(i,N);            if(re==false)                {                    break;                }        }        if(re==false)            printf("YES\n");        else            printf("NO\n");    }    return 0;}