题目链接:http://poj.org/problem?id=1986 题意:就是老问题求val[u]+val[v]-2*val[root]就行.还有这题没有给出不联通怎么输出那么题目给出的数据一定 是联通的. 题解:就是单纯的lca. #include <iostream> #include <cstring> #include <vector> #include <cstdio> using namespace std; const int M = 8e…

题目链接: Distance Queries Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 11531   Accepted: 4068 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 l…

解题关键:LCA模板题 复杂度:$O(n\log n)$ #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #include<iostream> typedef long lon…

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>…

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…

题目链接:http://poj.org/problem?id=1986 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 prob…

题目链接:http://poj.org/problem?id=1986 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…

Distance Queries [题目链接]Distance Queries [题目类型]LCA Tarjan法 &题意: 输入n和m,表示n个点m条边,下面m行是边的信息,两端点和权,后面的那个字母无视掉,没用的.接着k,下面k个询问lca,输出即可 &题解: 首先看的这个 http://www.cnblogs.com/JVxie/p/4854719.html 大致懂了方法,之后又找了这个代码 http://blog.csdn.net/lianai911/article/details…

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…

Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计,即在购买之前判断余额.如果购买一个商品之前,卡上的剩余金额大于或等于5元,就一定可以购买成功(即使购买后卡上余额为负),否则无法购买(即使金额足够).所以大家都希望尽量使卡上的余额最少.某天,食堂中有n种菜出售,每种菜可购买一次.已知每种菜的价格以及卡上的余额,问最少可使卡上的余额为多少.   Input 多组数据.对于每组数据:第一行为正整数n,表示菜的数量.n<=1000.第二行包括n个正整数,表示每种菜的价格…