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…

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…

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…

题目如下: 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 r…

problem 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…

For example, if we have 3 cities and 2 highways connecting city1-city2 and city1-city3. Then if city1 is occupied by the enemy, we must have 1 highway repaired, that is the highway city2-city3. Input Each input file contains one test case. Each case…

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…

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 of…

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…

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…

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…

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) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 cit…

1013. Battle Over Cities (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 cit…

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 othe…

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…

PAT甲级1013. Battle Over Cities 题意: 将所有城市连接起来的公路在战争中是非常重要的.如果一个城市被敌人占领,所有从这个城市的高速公路都是关闭的.我们必须立即知道,如果我们需要修理任何其他高速公路,以保持其他城市的连接.鉴于所有其余高速公路标记的城市地图, 你应该告诉高速公路需要修理的次数很快. 例如,如果我们有3个城市和2个连接city1-city2和city1-city3的高速公路3.那么如果city1被敌人占领,那么我们必须有1条公路修好,那就是高速公路city…

PAT甲级 1013 Battle Over Cities C++ 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…

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…

这题用并查集或者dfs都可以做 dfs #include<bits/stdc++.h> using namespace std; ; bool mp[N][N]; int n,m,k; bool vis[N]; void dfs(int v) { vis[v]=true; ;i<=n;i++){ if(mp[v][i]&&!vis[i]){ dfs(i); } } } int main() { fill(mp[],mp[]+N*N,false); scanf("…

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 of…

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 of…

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 of…

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 of…

好久都没有做题了,从长沙回来之后一直就是看看QT,感觉自己真的要蠢死了><不开心不开心 题目大概意思就是从一个图里面去掉一个点,看看剩下多少个孤立点. 自己想了好大一会儿没有思路,看到网上一个代码,真是惊叹好神奇...>< 用遍历的方式,如DFS,将去掉的点设为1,然后遍历一次看看剩下多少个没有被遍历到的点. #include <iostream> #include <cstring> using namespace std; #define MAX_VER…

这道题目的意思是:在战争时代,如果一个城市被敌人占领了,那么和该城市相连的道路都必须关闭,我们必须把剩下的城市(即不包括被敌人占领的城市)连接起来. 举个例子,我们有3个城市,C1,C2,C3,C1和C2相连,C1和C3相连.如果C1被敌人占领了,那么C2,C3就孤立了,C1是敌人的城市了,C2,C3是我们的城市,那么在C1被敌人占领的情况下,我们需要修一条路,即C2---C3之间连接起来…

使用一个标记数组,标记 节点是否已访问 int 连通度=0 dfs(node i) {标记当前节点为以访问 for(每一个节点) {if(当前几点未访问 并且 从i到当前节点有直接路径) dfs(当前节点) }} main() { .... for(对于每个点) {如果mark[i]==false://未被访问 {连通度++:dfs(i):} ... } 对于不管是任何带有循环性质的结构(dfs ,bfs,while,for) 由于边界问题,如果处理不当,会给思路和编码带来巨大的困难 一种解决方…

一.技术总结 这一题是考查图的知识,题目的意思要理解清楚,就是考查统计图中连通块的数量,也就是没有一个结点后. 怎么删除该结点,并且统计连通块的数量成为问题解决的关键,这里可以当访问到结点时,直接返回,或则跳过,这种操作就是相当于删除了该结点. memset(inq, false, sizeof(inq));这个是初始化标记数组,可以用于反复的遍历数组. 还有这次出现了一个比较简单的问题,就是在写for循环的嵌套时,同时使用了i变量,导致答案错误,这种问题一下又检查不出来,细心点. 二.参考代码…