题目描述
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.
John在他的农场中闲逛时发现了许多虫洞。虫洞可以看作一条十分奇特的有向边，并可以使你返回到过去的一个时刻（相对你进入虫洞之前）。John的每个农场有M条小路（无向边)连接着N （从1..N标号）块地，并有W个虫洞。其中1<=N<=500,1<=M<=2500,1<=W<=200。 现在John想借助这些虫洞来回到过去（出发时刻之前），请你告诉他能办到吗。 John将向你提供F(1<=F<=5)个农场的地图。没有小路会耗费你超过10000秒的时间，当然也没有虫洞回帮你回到超过10000秒以前。
输入输出格式
输入格式：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.
输出格式：Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).
输入输出样例
说明
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.
题目的大意应该已经很清楚了。
不过在此之前我也没做过负环的题，然后就按着最短路的思路写了一下。
最初我是直接跑一遍BFS版的SPFA，然后记录下遍历过的点。
如果某个点遍历过2次，那么就有环，否则无环。
然后一遍过了样例就直接交上去了。然后不加优化果然是T了的。
然后某位dalao告诉我负环可以用DFS版的SPFA。
其实两个版本的SPFA都差不多。
思路和注释差不多都在代码里了 其实是我好像没什么好讲的了233。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAXN=10001;
int k,n,m,w,tot=0;
int Head[MAXN],Dis[MAXN];//Head表示邻接表的头点，Dis[i]表示从起始点到i点的当前距离。
bool visit[MAXN],flag=0;//visit记录是否遍历到了这个点。
struct edge //邻接表存图
{
int next,node,w;
}h[MAXN*4];
void add(int u,int v,int w)//加点
{
h[++tot].next=Head[u];
h[tot].node=v;
h[tot].w=w;
Head[u]=tot; //虽然有重边但是不用去判断了。
}
void SPFA(int x)
{
if(flag)
return ;
visit[x]=1; //记录下所有遍历的点
for(int i=Head[x];i;i=h[i].next) //和跑最短路基本一样
{
if(flag)
return ;
int v=h[i].node;
if(Dis[v]>Dis[x]+h[i].w)
{
Dis[v]=Dis[x]+h[i].w;
if(visit[v]) //如果已经遍历过了就直接跳出去了。
{
flag=1;
return ;
}
else
SPFA(v); //继续遍历
}
}
visit[x]=0; //回溯的时候将遍历过的点还原
}
int main()
{
cin>>k;
while(k--)
{
tot=0;//每个数据要记得清0,
flag=0;
memset(h,0,sizeof(h));
memset(visit,0,sizeof(visit));
memset(Head,0,sizeof(Head));
memset(Dis,0,sizeof(Dis));//这里需要注意一下，Dis初始设置成0能跳过很多路径，
cin>>n>>m>>w; //直接找到负数点所在的环。
for(int i=1;i<=m;i++)
{
int x,y,z;
cin>>x>>y>>z;
add(x,y,z);
add(y,x,z); //正权边是双向的需要注意一下。
}
for(int i=1;i<=w;i++)
{
int x,y,z;
cin>>x>>y>>z;
add(x,y,-z); //这道题很方便的就是负权边都放一起了不用去打判断。
}
for(int i=1;i<=n;i++) //其实吧，这里随便找个点SPFA都行，毕竟有负环从哪个点开始都能找到，
{ //而且只要跑一遍就行了。
if(flag)
break; //发现负环直接退出循环。
SPFA(i);
}
if(flag)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}