PAT 1013 Battle Over Cities DFS深搜】的更多相关文章

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…
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) 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…
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…
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 ot…
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…
[编程题] 黑白树 时间限制:1秒 空间限制:32768K 一棵n个点的有根树,1号点为根,相邻的两个节点之间的距离为1.树上每个节点i对应一个值k[i].每个点都有一个颜色,初始的时候所有点都是白色的. 你需要通过一系列操作使得最终每个点变成黑色.每次操作需要选择一个节点i,i必须是白色的,然后i到根的链上(包括节点i与根)所有与节点i距离小于k[i]的点都会变黑,已经是黑的点保持为黑.问最少使用几次操作能把整棵树变黑. 输入描述: 第一行一个整数n (1 ≤ n ≤ 10^5) 接下来n-1…
DFS 深搜专题 入门典例 -- 凌宸1642 深度优先搜索 是一种 枚举所有完整路径以遍历所有情况的搜索方法 ,使用 递归 可以很好的实现 深度优先搜索. 1 最大价值 题目描述 ​ 有 n 件物品,每件物品的重量为 w[i] , 价值为 c[i] .现在需要选出若干件物品放入一个容器为 V 的背包中,使得在选入背包的物品重量和不超过容量 V 的前提下 ,让背包中的物品的价值之和最大,求最大价值.(1 ≤ n≤ 20 ) 输入描述: ​ 第一行输入物品总数 n 和 背包容量 v . ​ 第二行…
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…
[题目链接:HDOJ-2952] Counting Sheep Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2476    Accepted Submission(s): 1621 Problem Description A while ago I had trouble sleeping. I used to lie awake,…
深搜,从一点向各处搜找到全部能走的地方. Problem Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on…
Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only…
Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18201   Accepted: 9192 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100…
 Stamps  The government of Nova Mareterrania requires that various legal documents have stamps attached to them so that the government can derive revenue from them. In terms of recent legislation, each class of document is limited in the number of st…
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…
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…
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…
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…
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…
题目 In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or equal to (in a mi…
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…
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…
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…