题意 题目链接 问从\(0\)出发能否回到\(0\)且边权为负 Sol 先用某B姓算法找到负环,再判一下负环上的点能否到\(0\) #include<bits/stdc++.h> #define chmax(x, y) (x = (x > y ? x : y)) #define chmin(x, y) (x = (x < y ? x : y)) #define Pair pair<int, int> #define MP make_pair #define fi fir…

Sightseeing Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9703   Accepted: 3299 Description Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to…

Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 46123 Accepted: 17033 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…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24249   Accepted: 8652 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 pa…

Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 42366 Accepted: 15560 传送门 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 p…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 36425   Accepted: 13320 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 p…

链接: http://poj.org/problem?id=3259 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/B Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 25079   Accepted: 8946 Description While exploring his many farms, Farmer…

题目链接:http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 55082   Accepted: 20543 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is ver…

https://vjudge.net/problem/UVA-11090 平均权值最小的回路 为后面的做个铺垫 二分最小值,每条边权减去他,有负环说明有的回路平均权值小于他 spfa求负环的时候可以先把所有点加到队列里,d[i]=0 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; ,INF=1e9; inl…

这道题普通的bfs spfa或者ballen ford会T 所以我们使用dfs spfa 原因在于,bfs sfpa中每个节点的入队次数不定,退出操作不及时,而dfs则不会 既然,我们需要找负环,那么我们不妨将dis数组初始化为0,以每个点为起点进行dfs spfa 这样第一次扩展到的只有边权为负的边,之后若再次走到以访问过的点一定是负权回路 记得每次更换起点时清零vis数组 #include<cstdio> #include<cstring> #include<algori…