poj 3259 Wormholes ([kuangbin带你飞]专题四 最短路练习)

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.

题目大意：

农夫 FJ 有 N 块田地【编号 1...n】 (1<=N<=500)

        田地间有 M 条路径 【双向】(1<= M <= 2500)

        同时有 W 个孔洞,可以回到以前的一个时间点【单向】(1<= W <=200)

        问：FJ 是否能在田地中遇到以前的自己

算法：flod算法

思路： 田地间的双向路径加边,权值为正
       
孔洞间的单向路径加边,权值为负【可以回到以前】
       
判断有向图是否存在负环
     
因为如果存在了负数环,时间就会不停的减少,
     
那么 FJ 就可以回到以前更远的地方,肯定能遇到以前的自己的

<pre name="code" class="html">#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn = 510;
const int inf = 9999999;
struct Node{
int u,v;
int t;
}E[2500*2+200+10];
int d[maxn];
int c;
int n,m,w;
bool flod(){
for(int i=1;i<maxn;i++) d[i]=inf;
d[1]=0; //到某个点的时间
for(int i=1;i<n;i++){
bool flag = false;

for(int j=0;j<c;j++){
int u=E[j].u;
int v=E[j].v;
int t=E[j].t;
if(d[v]>d[u]+t){ //g跟新每个点时间
d[v]=d[u]+t;
flag = true;
}
}
if(!flag) return false; //如果不能跟新则不存在负边
}
for(int i=0;i<c;i++){
if(d[E[i].v]>d[E[i].u]+E[i].t) return true;
}
return false;
}

int main(){
int f;
scanf("%d",&f);
while(f--){

scanf("%d%d%d",&n,&m,&w);
c=0;
for(int i=0;i<m;i++){ //m条双向正边
int u,v,t;
scanf("%d%d%d",&u,&v,&t);
E[c].u=u;
E[c].v=v;
E[c].t=t;
c++;
E[c].u=v;
E[c].v=u;
E[c].t=t;
c++;
}
for(int i=0;i<w;i++){ //w条单向负边
int u,v,t;
scanf("%d%d%d",&u,&v,&t);
E[c].u=u;
E[c].v=v;
E[c].t=-t;
c++;
}
if(flod()){
printf("YES\n");
}
else printf("NO\n");
}
return 0;
}