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…
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…
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…
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…
<题目链接> 题目大意:给你一棵树,然后进行q次询问,然后要你统计这q次询问中指定的两个节点最近公共祖先出现的次数. 解题分析:LCA模板题,下面用的是离线Tarjan来解决.并且为了代码的简洁,本代码用的是vector存图. #include<iostream> #include<cstdio> #include<cstring> #include<vector> using namespace std; ; vector<int>…
<题目链接> 题目大意: 给出一棵树,问任意两个点的最近公共祖先的编号. 解题分析:LCA模板题,下面用的是树上倍增求解. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; struct Edge…
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…
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…
Distance Queries   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 t…
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…
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…
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…
好像是模板题  当作练习题 不错:  要求任意两点之间的距离.可以假设一个根节点,然后所有点到根节点的距离,然后求出任意两点多公共祖先:  距离就变成了 dis[u]+dis[v] - 2*dis[ lca(u,v) ]  非常好的题目 #include<iostream> #include<stdio.h> #include<cstring> #include<algorithm> #include<cmath> using namespace…
Distance Queries Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 14392   Accepted: 5066 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…
本题目输入格式同1984,这里的数据范围坑死我了!!!1984上的题目说边数m的范围40000,因为双向边,我开了80000+的大小,却RE.后来果断尝试下开了400000的大小,AC.题意:给出n个点,m条边. 接下来m行,每行对应u,v,len,字符(字符表示v位于u的哪个方向,在本题没有丝毫用处) 表示u和v的路径长度为len.注意本题是双向边! 给出k个查询,让你求两点间的距离.思路:可以这样处理:先处理出每个节点i到根的距离dist[i]. 设根节点1到a的路径和1到b的路径的最后一个…
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,…
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; ; ;//…
How far away ? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 25408    Accepted Submission(s): 10111 Problem Description There are n houses in the village and some bidirectional roads connectin…
http://poj.org/problem?id=1986 Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 14383   Accepted: 5063 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 lei…
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…
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…
题目链接:传送门 题意: 给定一棵树,求两个点之间的距离. 分析: LCA 的模板题目 ans = dis[u]+dis[v] - 2*dis[lca(u,v)]; 在线算法:详细解说 传送门 代码例如以下: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 40010; str…
传送门:http://poj.org/problem?id=1287 题意:给出n个点 m条边 ,求最小生成树的权 思路:最小生树的模板题,直接跑一遍kruskal即可 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<string.h> using namespace std; const int maxn = 5005; struct node { int u; in…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2586 题目大意:在一个无向树上,求一条链权和. 解题思路: 0 | 1 /   \ 2      3 设dist[i]为i到根0的链和,求法(Dfs过程中dist[v]=dist[u]+e[i].w) 对于树中任意两点形成的链,可以通过LCA最近公共祖先剖分. 比如2->3,就可以经过LCA点1:  2->1->3 链和=dist[u]+dist[v]-2*dist[LCA[u,v]] (…
题目: BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor, Jack Swigert, has asked her to benchmark…
<题目链接> 题目大意: 给定一段序列,进行q次询问,输出每次询问区间的最大值与最小值之差. 解题分析: RMQ模板题,用ST表求解,ST表用了倍增的原理. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long ll; ; int n,q; ll maxsum[M][],mi…
Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16748   Accepted: 6522 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled…
"Shortest" pair of paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1215   Accepted: 491 Description A chemical company has an unusual shortest path problem. There are N depots (vertices) where chemicals can be stored. There ar…
给出一棵树,对于每一个询问,给出2个节点,输出2个节点的距离. 输入中有字母,那个是没有用的,不用管. 思路: 0.选择编号为1的节点作为树的root (注意:有些题的边是单向的,这时候我们要根据节点的入度来确定root, 双向的话一般可以随意选择一个节点作为root) 1.dfs1,求出dep和pa[i][0] 2.初始化数组pa 3.节点(u,v)的权值为w 把本来是边的权值w赋给u,v中dep较大的节点, cost[i]表示节点i的权值为cost[i] 先初始化:cost[root]=0…