Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4425    Accepted Submission(s): 1263 Problem Description After World War X, a lot of cities have been seriously damag…

Connections between cities [题目链接]Connections between cities [题目类型]LCA Tarjan &题意: 输入一个森林,总节点不超过N(N<10000),由C次询问(C<1000000),每次询问两个点,如果来联通输出,两点之间的距离,如果不来联通,输出"Not connected" &题解: md,就没人吐槽这题询问时有相等的情况吗?我被这个坑了一天= = 最后把ans数组置为-1才过的,以前一直初…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 题意是给你n个点,m条边(无向),q个询问.接下来m行,每行两个点一个边权,而且这个图不能有环路.然后接下来q行,每行给你两个点,问你这两个点的最短距离是多少,要是不相连,则输出一串英文. 首先想到的是用(二分)倍增LCA,但是这题的坑点是两个点可能不在同一个图中,所以我dfs的时候用block[i]标记这个点属于哪一个图中,要是这个点在同一个图中,答案就是cost[u] + cost[v]…

题目链接 Connections between cities LCA的模板题啦. #include <bits/stdc++.h> using namespace std; #define REP(i,n) for(int i(0); i < (n); ++i) #define rep(i,a,b) for(int i(a); i <= (b); ++i) #define dec(i,a,b) for(int i(a); i >= (b); --i) #define for…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some mat…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11964    Accepted Submission(s): 2786 Problem Description After World War X, a lot of cities have been seriously damag…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4057    Accepted Submission(s): 1178 Problem Description After World War X, a lot of cities have been seriously damage…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11927    Accepted Submission(s): 2775 Problem Description After World War X, a lot of cities have been seriously dama…

http://acm.hdu.edu.cn/showproblem.php?pid=2874 题意: 求两个城市之间的距离. 思路: LCA题,注意原图可能不连通. 如果不了解离线算法的话,可以看我之前博客写的解释http://www.cnblogs.com/zyb993963526/p/7295894.html #include<iostream> #include<algorithm> #include<cstring> #include<cstdio>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 题目大意:给出n个点,m条边,q个询问,每次询问(u,v)的最短距离,若(u,v)不连通即不在同一颗树上则输出“Not connected”. 解题思路:这题也是模板题,有所不同的是这次给出的是森林而不是一棵树,所以vis数组得稍作修改,标记vis数组的是当前树的编号.下面给出Tarjan和倍增法两种解法. Tarjan(离线)写法,被MLE坑了,离线写法必须要用静态邻接表,因为虽然n不大,但…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=2874 [题目大意] 有n个村庄,m条路,不存在环,有q个询问,问两个村庄是否可达, 如果可达则输出最短路. [题解] 因为不存在环,所以是森林,我们计算每个连通块的dfs序,计算块内每个点到根距离 当两个点在同一个连通块时,我们输出其dis值之和减去其LCA的dis值, 否则输出不想连. [代码] #include <cstdio> #include <algorithm> #in…

题意: 城市 道路  没有环 不一定连通的树 求两城市的最短距离 设想一下就是很多小树  那好办 思路: lca离线算法 然后有个技巧就是 每次我们tarjan一棵树不是最后树的节点都访问过并且孩子全能找到老爸嘛 那么我们只要做做做做 做到全部的城市都访问过了  就行了 反正你做这颗小树的时候又不会影响到其他树的 #include <cstdio> #include <cstring> #include <algorithm> using namespace std;…

题目链接: Connections between cities Time Limit: 10000/5000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others) Problem Description   After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However,…

Problem Description After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city.…

题意:n棵树,求任意两点的最短距离. 解题关键:并查集判断两点是否位于一棵树上,然后求最短距离即可.此题可以直接对全部区间直接进行st表,因为first数组会将连接的两点的区间表示出来. //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib&…

图不一定联通,所以用并查集找各个联通块的祖先分别建图,之后就和LCA的步骤差不多了 #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<algorithm> #include<vector> using namespace std; inline int read(){ int sum=0,x=1; char ch=getc…

Connections between cities Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 896 Accepted Submission(s): 236   Problem Description After World War X, a lot of cities have been seriously damaged, an…

Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1444    Accepted Submission(s): 329 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…

hdu 6203 ping ping ping(LCA+树状数组) 题意:给一棵树,有m条路径,问至少删除多少个点使得这些路径都不连通 \(1 <= n <= 1e4\) \(1 <= m <= 5e4\) 思路: 根据路径的LCA深度从大到小排序,每次选择一个没被删除的LCA删除 当某个点删除时,跨越了以这个点为根的子树的路径都会被割断,而排序保证在同一子树内部的路径已经被处理过了,子树信息可以用dfs序来表示,区间操作可以用左右端点打标记,用树状数组维护即可. #include…

dfs找出所有节点所在树及到树根的距离及深度及父亲. i和j在一棵树上,则最短路为dis[i]+dis[j]-dis[LCA(i,j)]*2. #include <cstring> #include <cstdio> #define N 10005 #define add(u,v,w) e[++cnt]=(edge){v,head[u],w};head[u]=cnt using namespace std; struct edge{ int to,next,w; }e[N<&…

题意: 给出n个点m条边的图,c次询问 求询问中两个点间的最短距离. 解析: Floyd会T,所以用到了最短路树..具体思想为: 设k为u和v的最近公共祖先 d[i] 为祖结点到i的最短距离  则dis[u][v] = d[u] + d[v] - 2*d[k] 用tarjan的lca求即可 把这题代码当作模板就好啦 #include <iostream> #include <cstdio> #include <sstream> #include <cstring&…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 题目: Problem Description After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 这题有不连通的情况,特别注意. 觉得是存query的姿势不对,用前向星存了一遍,还是T…… /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏┓┃キリキリ♂ mind! ┛┗┛┗┛┃＼○/ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┃┃┃┃┃┃ ┻┻┻┻┻┻ */ #include <algorithm>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 思路:近乎纯裸的LCA,只是题目给出的是森林,就要判断是否都在同一颗树上,这里我们只需判断两个子节点的根节点是否相同就可以了. http://paste.ubuntu.com/5957639/…

传送门:Problem 2874 https://www.cnblogs.com/violet-acmer/p/9686774.html 改了一晚上bug,悲伤辣么大,明天再补详细题解 题解: 题目中说了无环的 也就是题目给的是一棵树或者是一片森林 通过加虚点可以把森林转化成为树 那么就是求树上任意2点之间的距离 2点之间的距离就是最近公共祖先分别这2个点的距离和 那么这个问题就被转化成了LCA问题. (以上参考资料:http://blog.sina.com.cn/s/blog_70743a56…

给出N个点,M条边.Q次询问 Q次询问每两点之间的最短距离 典型LCA 问题   Marjan算法解 #include "stdio.h" #include "string.h" struct Edge { int to,next,len; }edge[20010]; struct Ques { int to,next,index; }ques[2000010]; int head[10010],q_head[10010],f[10010],dis[10010];…

题目链接 给n个城市, m条边, q个询问, 每个询问, 输出城市a和b的最短距离, 如果不联通, 输出not connected. 用并查集判联通, 如果不连通, 那么两个联通块之间加一条权值很大的边. 然后树链剖分..... #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <c…

[HDU]P2586 How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 18675    Accepted Submission(s): 7274 Problem Description There are n houses in the village and some bidirectional road…

http://acm.hdu.edu.cn/showproblem.php?pid=3078 题意:给出n个点n-1条边m个询问,每个点有个权值,询问中有k,u,v,当k = 0的情况是将u的权值修改成v,当k不为0的情况是问u和v的路径中权值第k大的点的权值是多少. 思路:比较暴力的方法,可能数据太水勉强混过去了.对于每一个询问的时候保留两个点之间的lca,还有计算出两个点之间的点的个数(询问的时候如果点的个数小于k就不用算了),然后tarjan算完之后对每个询问再暴力路径上的每个点放进vec…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3371 Connect the Cities Description In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The…