poj 1986 Distance Queries(LCA)】的更多相关文章

Description Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in…
计算树上的路径长度.input要去查poj 1984. 任意建一棵树,利用树形结构,将问题转化为u,v,lca(u,v)三个点到根的距离.输出d[u]+d[v]-2*d[lca(u,v)]. 倍增求解: #include<cstdio> #include<cstring> #include<queue> #include<algorithm> #define rep(i,a,b) for(int i=a;i<=b;i++) #define clr(a,…
Distance Queries Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 12846   Accepted: 4552 Case Time Limit: 1000MS Description Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifesty…
题目链接 Description Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input…
Distance Queries Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 12950   Accepted: 4577 Case Time Limit: 1000MS Description Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifesty…
1.一颗树中,给出a,b,求最近的距离.(我没考虑不联通的情况,即不是一颗树的情况) 2.用最近公共祖先来求, 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<string.h> using namespace std; ; ;//…
POJ 1986 Distance Queries / UESTC 256 Distance Queries / CJOJ 1129 [USACO]距离咨询(最近公共祖先) Description Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path…
POJ.1986 Distance Queries ( LCA 倍增 ) 题意分析 给出一个N个点,M条边的信息(u,v,w),表示树上u-v有一条边,边权为w,接下来有k个询问,每个询问为(a,b),求a,b两点到lca(a,b)的边权之和为多少. 倍增维护树上前缀和,求得LCA之后,相应做差即可. 代码总览 #include <cstdio> #include <algorithm> #include <cstring> #include <cmath>…
标题来源:POJ 1986 Distance Queries 意甲冠军:给你一棵树 q第二次查询 每次你问两个点之间的距离 思路:对于2点 u v dis(u,v) = dis(root,u) + dis(root,v) - 2*dis(roor,LCA(u,v)) 求近期公共祖先和dis数组 #include <cstdio> #include <cstring> #include <vector> using namespace std; const int max…
任意门:http://poj.org/problem?id=1986 Distance Queries Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 16648   Accepted: 5817 Case Time Limit: 1000MS Description Farmer John's cows refused to run in his marathon since he chose a path much t…