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;
}