开两个集合,一个存储当前顶点可以到达的点,另一个存储当前顶点不能到达的点.如果可以到达,那肯定由该顶点到达是最短的,如果不能,那就留着下一次再判. #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef pair<int,int> pll; const int INF = 0x3f3f3f3f; +; int T; int n, m, s; int d[maxn]; vector<i…

---恢复内容开始--- Sparse Graph Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2590    Accepted Submission(s): 902 Problem Description In graph theory, the complement of a graph G is a graph H on t…

题目:这里 题意: 相当于一开始给一个初始好了的无向完全图给你,然后给让你删除m条边,再给你一个点v,最后问你在剩下的图里从这个点v出发能到达所有边点的最小路径是多少? 一看是所有点的最小路径,一看就觉得是个bfs,记忆化搜一下然后加个优化什么的,由于数据不知道是个什么奇葩而且比赛中还改数据,所以很多人wa的莫名其妙, 过也过的莫名其妙,我虽然过了但觉得有点不靠谱,赛后看了https://async.icpc-camp.org/d/546-2016的题解思路写了一发,总感觉更靠谱一点. 之前自己…

Sparse Graph Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 689    Accepted Submission(s): 238 Problem Description In graph theory, the complement of a graph G is a graph H on the same vertic…

1.HDU 5876  Sparse Graph 2.总结:好题,把STL都过了一遍 题意:n个点组成的完全图,删去m条边,求点s到其余n-1个点的最短距离. 思路:把点分为两个集合,A为所有没有到达过的点,B为当前不可到达的点,每次拓展A中可到过的点加入队列. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #inc…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5876 题意:给定一个图(n个顶点m条边),求其补图最短路 思路:集合a表示当前还未寻找到的点,集合b表示本次bfs之后仍未寻找到的点 #include<cstdio> #include<set> #include<queue> #include<cstring> using namespace std; const int N = 2e5 + 5; set &l…

Hunter Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2014    Accepted Submission(s): 615 Problem Description One day, a hunter named James went to a mysterious area to find the treasures. Jame…

Travel The country frog lives in has \(n\) towns which are conveniently numbered by \(1, 2, \dots, n\). Among \(\frac{n(n - 1)}{2}\) pairs of towns, \(m\) of them are connected by bidirectional highway, which needs \(a\) minutes to travel. The other…

http://acm.hdu.edu.cn/showproblem.php?pid=5876 题意: 在补图中求s到其余各个点的最短路. 思路:因为这道题目每条边的距离都是1,所以可以直接用bfs来做. 处理的方法是开两个集合,一个存储当前顶点可以到达的点,另一个存储当前顶点不能到达的点.如果可以到达,那肯定由该顶点到达是最短的,如果不能,那就留着下一次再判. #include<iostream> #include<algorithm> #include<cstring>…

In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are not adjacent in G. Now you are given an undirected graph G of N nodes and M bidirectional edges o…