Wormholes（bellman-Ford）

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.

一个地图，有虫洞（可以回到过去），问有没有可能，从一个地方出发，通过虫洞，能在自己进虫洞之前看到自己。

其实说白了就是找负环，直接套模板（这里的套模板指的是套思路，而不是copy代码，模板手写更容易理解，并且这中模板一点也不复杂，没有copy必要）

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int head[505],d[505];
struct Node
{
int u,v,w,next;
}node[5010];
void add_edge(int u,int v,int w,int tot)
{
node[tot].u = u;node[tot].v = v;
node[tot].w = w;
node[tot].next = head[u];
head[u] = tot;
}
void relax(int u,int v,int w)
{
d[v] = min(d[v],d[u] + w);
}
int main()
{
#ifdef LOCAL
freopen("C:\\Users\\巍巍\\Desktop\\in.txt","r",stdin);
//freopen("C:\\Users\\巍巍\\Desktop\\out.txt","w",stdout);
#endif // LOCAL
int t;
scanf("%d",&t);
while(t--)
{
int n,m,w;
scanf("%d%d%d",&n,&m,&w);
for(int i = 1;i <= n;i++)head[i] = -1;
for(int i = 1;i <= m;i++)
{
int from,to,wei;
scanf("%d%d%d",&from,&to,&wei);
add_edge(from,to,wei,2*i - 1);
add_edge(to,from,wei,2*i);
}
for(int i = 1;i <= w;i++)
{
int from,to,wei;
scanf("%d%d%d",&from,&to,&wei);
add_edge(from,to,-wei,2*m + i);
}
for(int i = 1;i <= n;i++)d[i] = inf;
d[1] = 0;
for(int i = 1;i < n;i++)
for(int j = 1;j <= 2*m + w;j++)
relax(node[j].u,node[j].v,node[j].w);
bool flag = 0;
for(int i = 1;i <= 2*m + w;i++)
{
int from = node[i].u;
int to = node[i].v;
if(d[to] > d[from] + node[i].w)
{
flag = 1;break;
}
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}