F - Wormholes
题目链接：https://vjudge.net/contest/66569#problem/F
题目：
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：
农夫约翰在探索他的许多农场，发现了一些惊人的虫洞。虫洞是很奇特的，因为它是一个单向通道，可让你进入虫洞的前达到目的地！他的N（1≤N≤500）个农场被编号为1..N，之间有M（1≤M≤2500）条路径，W（1≤W≤200）个虫洞。FJ作为一个狂热的时间旅行的爱好者，他要做到以下几点：开始在一个区域，通过一些路径和虫洞旅行，他要回到最开时出发的那个区域出发前的时间。也许他就能遇到自己了:)。为了帮助FJ找出这是否是可以或不可以，他会为你提供F个农场的完整的映射到（1≤F≤5）。所有的路径所花时间都不大于10000秒，所有的虫洞都不大于万秒的时间回溯。
Input
第1行：一个整数F表示接下来会有F个农场说明。 每个农场第一行：分别是三个空格隔开的整数：N，M和W 第2行到M+1行：三个空格分开的数字（S，E，T）描述，分别为：需要T秒走过S和E之间的双向路径。两个区域可能由一个以上的路径来连接。 第M +2到M+ W+1行：三个空格分开的数字（S，E，T）描述虫洞，描述单向路径，S到E且回溯T秒。
Output
F行，每行代表一个农场 每个农场单独的一行，” YES”表示能满足要求，”NO”表示不能满足要求。
思路：
套判断负环的模板即可，spfa算法如下：
//
// Created by hanyu on 2019/7/19.
//
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=2e5+7;
#define MAX 0x3f3f3f3f
int book[maxn],cnt[maxn],d[maxn],head[maxn];
int n,m,w;
int pos;
struct Node{
int s;
int e;
int t;
}node[maxn];
void add(int s,int e,int t)
{
node[pos].s=e;
node[pos].e=t;
node[pos].t=head[s];
head[s]=pos++;
}
bool spfa(int start)
{
queue<int>qu;
qu.push(start);
book[start]=1;
d[start]=0;
while(!qu.empty())
{
int now=qu.front();
qu.pop();
book[now]=0;
for(int i=head[now];i!=-1;i=node[i].t)
{
int ss=node[i].s;
int ee=node[i].e;
if(d[ss]>d[now]+ee)
{
d[ss]=d[now]+ee;
if(!book[ss])
{
qu.push(ss);
book[ss]=1;
cnt[ss]++;
if(cnt[ss]>=n)
return true;//判断负环
}
}
}
}
return false;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(book,0,sizeof(book));
memset(cnt,0,sizeof(cnt));
memset(d,MAX,sizeof(d));
memset(head,-1,sizeof(head));
memset(node,0,sizeof(node));
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);
add(s,e,t);
add(e,s,t);
}
for(int i=1;i<=w;i++)
{
scanf("%d%d%d",&s,&e,&t);
add(s,e,-t);
}
cnt[1]=1;
pos=0;
if(spfa(1))
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
原文地址：https://www.cnblogs.com/Vampire6/p/11216351.html