HDU - 2586 How far away ? Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like th…

hdu 2586 How far away ?倍增LCA 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2586 思路: 针对询问次数多的时候,采取倍增求取LCA,同时跟新距离数组 因为 \(2^{16} > 40000\) 所以所以表示祖先的数组dp[][]第二维取到16即可 就这道题来说,与比较tarjan比较,稍快一点 代码: #include <iostream> #include <algorithm> #includ…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=2586 How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 24439    Accepted Submission(s): 9739 Problem Description There are n houses in…

2586.How far away ? 这个题以前写过在线LCA(ST)的,HDU2586.How far away ?-在线LCA(ST) 现在贴一个离线Tarjan版的 代码: //A-HDU2586-LCA-tarjan离线版 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<bitset> #include<casse…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 LCA模版题. RMQ+LCA: #include <iostream> #include <cstdio> #include <cstring> #include <vector> using namespace std; ; typedef pair<int , int>P; vector <P> G[MAXN]; ] , ok…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 最近公共祖先问题~~LAC离散算法 题目大意:一个村子里有n个房子,这n个房子用n-1条路连接起来,接下了有m次询问,每次询问两个房子a,b之间的距离是多少. 很明显的最近公共祖先问题,先建一棵树,然后求出每一点i到树根的距离dis[i],然后每次询问a,b之间的距离=dis[a]+dis[b]-2*dis[LCA(a,b)]; LCA(a,b)即是a,b的最近公共祖先.. 关于最近公共祖先,…

http://acm.hdu.edu.cn/showproblem.php?pid=2586 题意:给出一棵树,求出树上任意两点之间的距离. 思路: 这道题可以利用LCA来做,记录好每个点距离根结点的距离,然后只要计算出两点的LCA,这样一来答案就是distance[u]+distance[v]-2distance[LCA]. #include<iostream> #include<algorithm> #include<cstring> #include<cst…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 Problem Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2586 题意:给你N个点,M次询问.1~N-1行输入点与点之间的权值,之后M行输入两个点(a,b)之间的最近距离: 思路:设d[maxn]是当前点到根结点的距离,则答案就是 d[a]-d[lca(a,b)]+d[b]-d[lca(a,b)]: 接下来介绍LCA(Least Common Ancestors) 即最近公共祖先,是指在有根树中,找出某两个结点u和v最近的公共祖先. 常见解法一般有三种 这里讲…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 20971    Accepted Submission(s): 8245 Problem Description There are n houses in the village and some bidirectional roads connecting…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How f…

2018.9.10 0:40 重新敲一遍,然后很认真的写了注释,方便自己和队友看,刚过去的一天的下午打网络赛有一题用到了这个,但是没写注释,队友改板子有点伤,因为我没注释... 以后写博客,代码要写注释,要把题目意思写上... 我好捞啊... How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 21408    …

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9359    Accepted Submission(s): 3285 Problem Description There are n houses in the village and some bidirectional roads connecting…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8712    Accepted Submission(s): 3047 Problem Description There are n houses in the village and some bidirectional roads connecting…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17449    Accepted Submission(s): 6747 Problem Description There are n houses in the village and some bidirectional roads connecting…

#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<vector> #include<cmath> #include<map> using namespace std; ; struct N { int v,d; N(int x,int y) { v=x; d=y; } }; vector<N>g[…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18652    Accepted Submission(s): 7268 Problem Description There are n houses in the village and some bidirectional roads connecting them. Every da…

倍增求LCA LCA函数返回(u,v)两点的最近公共祖先 #include <bits/stdc++.h> using namespace std; *; struct node { int v,val,next; node(){} node(int vv,int va,int nn):v(vv),val(va),next(nn){} }E[N]; int n,m; ],dep[N]; void init() { tot = ; memset(head,,sizeof(head)); mems…

题意:给定N个节点一棵树,现在要求询问任意两点之间的简单路径的距离,其实也就是最短路径距离. 析:用LCA问题的Tarjan算法,利用并查集的优越性,产生把所有的点都储存下来,然后把所有的询问也储存下来,然后从树根开始搜索这棵树, 在搜索子树的时候,把并查集的父结点不断更新,在搜索时计算答案,d[i] 表示从根结点到 i 结点的距离,然后分别计算到两个结点的距离, 再减去2倍的根结点到他们公共最近的结点距离,就OK了.不过这个题有个坑人的地方,不用输出那个空行,否则就是PE. 因为这个题询问比较…

1.给定一棵树,每条边都有一定的权值,q次询问,每次询问某两点间的距离. 2.这样就可以用LCA来解,首先找到u, v 两点的lca,然后计算一下距离值就可以了. 这里的计算方法是,记下根结点到任意一点的距离dis[],这样ans = dis[u] + dis[v] - 2 * dis[lca(u, v)] 3. /* 离线算法,LCATarjan 复杂度O(n+Q); */ #include<iostream> #include<stdio.h> #include<stri…

Time limit 1000 ms Memory limit 32768 kB Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it…

在dfs的过程中维护三个数组: deep[i],表示i点在树中的深度: grand[x][i],表示x的第2^i个祖先的节点编号: dis[x][i],表示x到它2^i祖 #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<algorithm> #include<vector> using namespace std; co…

HDU 2586 How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11320    Accepted Submission(s): 4119 Problem Description   There are n houses in the village and some bidirectional roads…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2586 How far away ? Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 27977    Accepted Submission(s): 11218 Problem Description There are n houses in the village and some bidirectional roads connectin…

题目大意:求树上任意两点距离. 思路: dis[i]表示i到根的距离(手动选根),则u.v的距离=dis[u]+dis[v]-2*dis[lca(u,v)]. lca:u~v的dfs序列区间里,深度最小的节点即为u.v的lca(最近公共祖先),RMQ把它找到.   难点: 何为dfs序: 就是树上dfs依次遍历的节点编号的序列.它的特点是回溯时会[再次]经过非叶节点,非叶节点在dfs序中出现多次,且dfs序列个数>节点个数. 为什么要用dfs序: 因为u.v的lca只出现在u.v的dfs序间.比…

题意:一个村子有n个房子,他们用n-1条路连接起来,每两个房子之间的距离为w.有m次询问,每次询问房子a,b之间的距离是多少. 分析:近期公共祖先问题,建一棵树,求出每一点i到树根的距离d[i],每次询问a.b之间的距离=d[a]+d[b]-2*d[LCA(a,b)];LCA(a,b)是a,b的近期公共祖先. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include…

How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11699    Accepted Submission(s): 4300 Problem Description There are n houses in the village and some bidirectional roads connectin…

How far away ? Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2586 Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How f…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 在线版本: 在线方法的思路很简单,就是倍增.一遍dfs得到每个节点的父亲,以及每个点的深度.然后用dp得出每个节点向上跳2^k步到达的节点. 那么对于一个查询u,v,不妨设depth[u]>=depth[v],先让u向上跳depth[u]-depth[v]步,跳的方法就是直接用数字的二进制表示跳. 然后现在u和v都在同一深度上了,再二分找向上共同的祖先,就可以二分出lca了.复杂度nlogn预…