标签：wormholes wing possible starting move can 空格 single sel
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.
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 8Sample Output
NO YESHint
// // 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; }
[kuangbin带你飞]专题四 最短路练习 F - Wormholes （判断负环）
标签：wormholes wing possible starting move can 空格 single sel
原文地址：https://www.cnblogs.com/Vampire6/p/11216351.html
