| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 44174 | Accepted: 16235 |
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 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
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 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.
题目大意：
一个农民有农场，上面有一些虫洞和路，走虫洞可以回到 T秒前，而路就和平常的一样啦，需要花费时间走过。问该农民可不可能从某个点出发后回到该点，并且于出发时间之前?
直接上spfa，看了很多模板，只有这个模板的命名规范些，以后就照这个模板来做吧。
题目大意：
一个农民有农场，上面有一些虫洞和路，走虫洞可以回到 T秒前，而路就和平常的一样啦，需要花费时间走过。问该农民可不可能从某个点出发后回到该点，并且于出发时间之前?
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 520 ;
const int maxm = 5200;
struct edge
{
int to;
int val;
int next;
}e[maxm];
int len,head[maxn];
int dis[maxn];
int n,m,c;
bool spfa()
{
for(int i = 1;i<=n;i++)
dis[i]=inf;
bool vis[maxn] = {0};
int cnt[maxn] = {0};
//因为这里是找环，不是找到某点的最短路径，所以不用初始dis[start]=0;直接从1开始找。
int cur = 1; //当前点初始化
queue<int> q;
q.push(cur);
vis[cur] = true;
cnt[cur]=1;
dis[cur]=0; //以当前点的为始点的最短路径
while(!q.empty())
{
cur=q.front();
q.pop();
vis[cur]=false;//出队则取消标记
for(int i = head[cur];i!=-1;i=e[i].next)
{
int tto = e[i].to;
if( dis[tto]>dis[cur]+e[i].val )
{
dis[tto] = dis[cur]+e[i].val;
if(!vis[tto]){
cnt[tto]++;
vis[tto]=true;
q.push(tto);//新点进队
if(cnt[cur]>n)
return true;//如果进队超过n次，则存在负环
}
}
}
}
return false;
}
void add(int from,int to,int val){
e[len].to=to;
e[len].val=val;
e[len].next=head[from];
head[from]=len++;
}
int main(){
int T;
scanf("%d",&T);
while(T--){
memset(head,-1,sizeof(head));
len=0;
scanf("%d%d%d",&n,&m,&c);
for(int i = 0;i<m;i++){
int from,to,val;
scanf("%d%d%d",&from,&to,&val);
add(from,to,val);
add(to,from,val);
}
for(int i = 0;i<c;i++)
{
int from,to,val;
scanf("%d%d%d",&from,&to,&val);
add(from,to,-val);
}
if(spfa())
printf("YES");
else
printf("NO");
printf("
");
}
return 0;
}