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 backT 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.
题意：走一条路会花费时间 走虫洞会时间倒流 问能不能走回起点的时候时间倒流
思路：其实就是判断有没有环 有环的话
看题的时候看了半天不知道起点到底是哪一个点
感觉现在bellman写的还挺顺手的了
一个加边的addedge函数 一个判断松弛的relax函数
然后外面一个for循环遍历n-1次 里面的for循环松弛每一条边
再对每一条边判断能不能松弛 能就说明有环
有环其实就说明这个路径上有负的
不然干吗要不停的重复走？？？
只有可以不断减小才会形成环
WA了一发是因为加边的时候的cnt没有初始化
代码：
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<cstring>
#include<queue>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
int f, n, m, w, cnt;
struct edge{
int s, e, t;
}path[5205];
long long d[505];
void addedge(int s, int e, int t)
{
path[cnt].s = s;
path[cnt].e = e;
path[cnt].t = t;
cnt++;
}
bool relax(int j)
{
if(d[path[j].e] > d[path[j].s] + path[j].t){
d[path[j].e] = d[path[j].s] + path[j].t;
return true;
}
return false;
}
bool bellman(int sec)
{
memset(d, inf, sizeof(d));
d[sec] = 0;
for(int i = 0; i < n - 1; i++){
bool flag = false;
for(int j = 0; j < cnt; j++){
if(relax(j)) flag = true;
}
if(!flag) return false;
}
for(int i = 0; i < cnt; i++){
if(relax(i)) return true;
}
return false;
}
int main()
{
cin>>f;
while(f--){
cin>>n>>m>>w;
cnt = 0;
for(int i = 0; i < m; i++){
int a, b, c;
cin>>a>>b>>c;
addedge(a, b, c);
addedge(b, a, c);
}
for(int i = 0; i < w; i++){
int a, b, c;
cin>>a>>b>>c;
addedge(a, b, -c);
}
if(bellman(1))
cout<<"YES\n";
else
cout<<"NO\n";
}
return 0;
}