并查集判断连通性. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> using namespace std; ; struct Edge { int u,v; }e[maxn*maxn]; int n,m,k; int f[maxn]; int Find(int x) { if…

1013. Battle Over Cities (25) t is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other…

Battle Over Cities It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways…

题意: 输入三个整数N,M,K(N<=1000,第四个数据1e5<=M<=1e6).有1~N个城市,M条高速公路,K次询问,每次询问输入一个被敌军占领的城市,所有和该城市相连的高速公路全部不能使用,求增加多少条高速公路可以使剩下N-1个城市联通.(原本城市之间可能不联通,假设原本联通只能通过第0,4个数据). trick: 同一份代码交很多次,有几次会第4个点超时.(存疑)建议采用g++而不是clang++ AAAAAccepted code: #include<bits/stdc…

1013 Battle Over Cities (25 分)   It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any o…

1013 Battle Over Cities (25分)   It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any ot…

这题给定了一个图,我用DFS的思想,来求出在图中去掉某个点后还剩几个相互独立的区域(连通子图). 在DFS中,每遇到一个未访问的点,则对他进行深搜,把它能访问到的所有点标记为已访问.一共进行了多少次这样的搜索, 就是我们要求的独立区域的个数. #include <iostream> #include <fstream> #include <memory.h> using namespace std; const int maxNum = 1001; bool visit…

题目 It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest…

https://www.patest.cn/contests/pat-a-practise/1013 思路:并查集合并 #include<set> #include<map> #include<queue> #include<algorithm> #include<string> #include<string.h> using namespace std; int n;//number of city int m;//number…

题目就是求联通分支个数删除一个点,剩下联通分支个数为cnt,那么需要建立cnt-1边才能把这cnt个联通分支个数求出来怎么求联通分支个数呢可以用并查集,但并查集的话复杂度是O(m*logn*k)我这里用的是dfs,dfs的复杂度只要O((m+n)*k)这里k是指因为有k个点要查询,每个都要求一下删除后的联通分支数.题目没给定m的范围,所以如果m很大的话,dfs时间会比较小. for一遍1~n个点,每次从一个未标记的点u开始dfs,标记该dfs中访问过的点.u未标记过,说明之前dfs的时候没访问过…