链接：Wormholes
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
共F个农场（即F组测试数据），每个农场有N片田地（编号1 ~ N），M条道路，W个虫洞。
对于道路，可以从S到E，也可以从E到S（即双向通行），耗费T时间。
对于虫洞，仅能从S到E（即单向通行），同时穿越回T时间之前。
问是否能从某位置出发，通过某些道路和虫洞，最终在最初离开时间之前回到起始位置。（即回到初始位置，但时间倒退了）
分析
理解一下题意就可以发现该题是要判断负环（定道路路权为正，虫洞路权为负），当存在负环时就可以利用负环令时间一直倒退了，从而达到目的。
最短路算法 BF 和 SPFA 都可以判断有无源点可达的负环。
虽然题目没告知，不过这题的测试数据中貌似所有结点都在同一连通块（源点都可达），因为直接以1为源点判断一次就能AC了。（如果有多个连通块，那么就得每次用属于不同连通的结点多判断几次了）
SPFA：
#include<cstdio>
#include<climits>
#include<algorithm>
#include<queue>
#include<cstring>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=510;
int N,M,W;
int G[maxn][maxn],d[maxn],cnt[maxn];
bool inq[maxn];
bool SPFA(int s)
{
memset(d,0x3f,sizeof(d));memset(cnt,0,sizeof(cnt)); memset(inq,false,sizeof(inq));queue<int> q;d[s]=0;q.push(s);inq[s]=true;cnt[s]++;while(!q.empty()){
int u=q.front();q.pop();inq[u]=false;for(int v=1;v<=N;v++){
if(G[u][v]!=INF&&d[u]+G[u][v]<d[v]){
d[v]=d[u]+G[u][v];if(!inq[v]){
q.push(v);inq[v]=true;cnt[v]++;if(cnt[v]>N-1)return true; }}}}return false;
}
int main()
{
int T;scanf("%d",&T);while(T--){
memset(G,0x3f,sizeof(G));scanf("%d %d %d",&N,&M,&W);int S,E,T;for(int i=1;i<=M;i++){
scanf("%d %d %d",&S,&E,&T);G[S][E]=min(T,G[S][E]);G[E][S]=min(T,G[E][S]);}for(int i=1;i<=W;i++){
scanf("%d %d %d",&S,&E,&T);G[S][E]=-1*T;}if(SPFA(1))printf("YES\n");elseprintf("NO\n");}return 0;
}
BF：
#include<cstdio>
#include<climits>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstring>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=510;
struct edge
{
int u,v,w;edge(int S,int E,int T):u(S),v(E),w(T){
}
};
int N,M,W;
vector<edge> a;
int d[maxn];
bool BF(int s)
{
memset(d,0x3f,sizeof(d));d[s]=0;for(int i=1;i<=N-1;i++){
for(int j=0;j<a.size();j++){
int u=a[j].u,v=a[j].v,w=a[j].w;if(d[u]+w<d[v])d[v]=d[u]+w;}}for(int j=0;j<a.size();j++){
int u=a[j].u,v=a[j].v,w=a[j].w;if(d[u]+w<d[v])return true;}return false;
}
int main()
{
int F;scanf("%d",&F);while(F--){
a.clear();scanf("%d %d %d",&N,&M,&W);int S,E,T;for(int i=1;i<=M;i++){
scanf("%d %d %d",&S,&E,&T);a.push_back(edge(S,E,T));a.push_back(edge(E,S,T));}for(int i=1;i<=W;i++){
scanf("%d %d %d",&S,&E,&T);a.push_back(edge(S,E,-1*T));}if(BF(1))printf("YES\n");elseprintf("NO\n");}return 0;
}