意甲冠军:它使树m路径,当被问及选择尽可能多的路径,而这些路径不相交. 思考:贪心,比較忧伤.首先求一下每对路径的lca.依照lca的层数排序.在深一层的优先级高.那么就能够贪心了,每次选择层数最深的那一个,假设两个端点没有被标记,那么就选择这条路径,把lca的子树都标记掉. 代码: #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio>…

题目链接:hdu 4912 Paths on the tree 题目大意:给定一棵树,和若干个通道.要求尽量选出多的通道,而且两两通道不想交. 解题思路:用树链剖分求LCA,然后依据通道两端节点的LCA深度排序,从深度最大优先选.推断两个节点均没被标 记即为可选通道. 每次选完通道.将该通道LCA下面点所有标记. #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #includ…

题目链接 Paths on the tree 来源  2014 多校联合训练第5场 Problem B 题意就是给出m条树上的路径,让你求出可以同时选择的互不相交的路径最大数目. 我们先求出每一条路径(u, v)中u和v的LCA:w,按照路径的w的深度大小deep[w]对所有的路径排序. deep[w]越大,排在越前面. 然后从第一条路径开始一次处理,看c[u]和c[v]是否都没被标记过,如果都没被标记过则我们把这条路径选上,把答案加1. 同时标记以w为根的子树的节点为1,方便后续对c数组的查询…

先给个LCA模板 HDU 1330(LCA模板) #include <cstdio> #include <cstring> #define N 40005 struct Edge{ int x,y,d,ne; }; Edge e[N*],e2[N*]; int be[N],be2[N],all,all2,n,m; bool vis[N]; int fa[N]; ]; int dis[N]; void add(int x, int y, int d, Edge e[], int be…

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…

Paths on the tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 297    Accepted Submission(s): 93 Problem Description bobo has a tree, whose vertices are conveniently labeled by 1,2,-,n. Th…

F. Three Paths on a Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given an unweighted tree with nn vertices. Recall that a tree is a connected undirected graph without cycle…

Dylans loves tree view code#pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <vector> #include <queue> #include <…

lca的做法还是非常明显的.简单粗暴, 只是不是正解.假设树是长链就会跪,直接变成O(n).. 最后跑的也挺快,出题人还是挺阳光的.. 动态树的解法也是听别人说能ac的.预计就是放在splay上剖分一下,做法还是比較复杂的.,, 来一发lca: #include <stdio.h> #include <iostream> #include <algorithm> #include <sstream> #include <stdlib.h> #in…

题意及思路 说一下为什么按LCA深度从深到浅贪心是对的.我们可以直观感受一下,一条的路径会影响以这个lca为根的这颗树中的链,而深度越深,影响范围越小,所以先选影响范围小的路径. #include <bits/stdc++.h> using namespace std; const int maxn = 100010; int head[maxn], Next[maxn * 2], ver[maxn * 2], tot, f[maxn][20]; int deep[maxn], t; stru…