# poj3259:负环判定，SPFA

Wormholes
 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 41416 Accepted: 15226

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.

Source

1. 正权道路是双向的；
2. 负权道路是单向的；
3. 题目的数据范围貌似小了；
4. 初始化...(大神请无视)

O(nke)做法吧貌似，不太会算..

```/*
Source Code

Problem: 3259		User: aclolicon
Memory: 596K		Time: 1766MS
Language: C++		Result: Accepted
Source Code
*/
#include
#include
#define MAXN 40000
#define INF 0x3f3f3f3f
using namespace std;
struct Edge{
int to;
int next;
int w;
}edge[MAXN];
int cnt = 0;
int n, m, w;

void add(int s, int e, int t){
edge[cnt].w = t;
edge[cnt].to = e;
}
bool spfa(int s){
queue  q;
q.push(s);
int cc[MAXN], dis[MAXN];
bool vis[MAXN];
for (int i = 0; i < n + 1; i++){
cc[i] = 0;
dis[i] = INF;
vis[i] = 0;
}
dis[s] = 0;
bool flag = 0;
while(!q.empty()){
int x = q.front();
q.pop();
vis[x] = 0;
for (int i = head[x]; i != -1 && !flag; i = edge[i].next){
int to = edge[i].to;
if (dis[x] + edge[i].w < dis[to]){
dis[to] = dis[x] + edge[i].w;
if (!vis[to]){
vis[to] = 1;
q.push(to);
cc[to]++;
if (cc[to] >= n){
flag = 1;
break;
}
}
}
}
if (flag) break;
}
if (flag) return 0;
return 1;
}

void init(){
for (int i = 0; i < n + 1; i++){
edge[i].to = -1;
edge[i].w = 0;
edge[i].next = -1;
}
}

void solve(){
bool flag = 0;
cnt = 0;
int s, e, t;
for (int i = 0; i < m; i++){
scanf("%d%d%d", &s, &e, &t);
}
for (int i = 0; i < w; i++){
scanf("%d%d%d", &s, &e, &t);
}
for (int i = 1; i <= n; i++){
if (!spfa(i)) {
flag = 1;
break;
}
}
if (flag) printf("YESn");
else printf("NOn");
}

int main(){
int f;
scanf("%d", &f);
while(f--) {
scanf("%d%d%d", &n, &m, &w);
init();
solve();
}
return 0;
}

```