Binary Apple Tree Time Limit: 1000ms Memory Limit: 16384KB This problem will be judged on Ural. Original ID: 101864-bit integer IO format: %lld      Java class name: (Any)     Let's imagine how apple tree looks in binary computer world. You're right,…

Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enumerate by integers the root of binary apple tree, points of branch…

CJOJ 1976 二叉苹果树 / URAL 1018 Binary Apple Tree(树型动态规划) Description 有一棵苹果树,如果树枝有分叉,一定是分2叉(就是说没有只有1个儿子的结点)这棵树共有N个结点(叶子点或者树枝分叉点),编号为1-N,树根编号一定是1.我们用一根树枝两端连接的结点的编号来描述一根树枝的位置.现在这颗树枝条太多了,需要剪枝.但是一些树枝上长有苹果. 给定需要保留的树枝数量,求出最多能留住多少苹果.下面是一颗有 4 个树枝的树. 2 5 \ / 3 4…

1018. Binary Apple Tree Time limit: 1.0 secondMemory limit: 64 MB Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enu…

http://vjudge.net/problem/17662 loli蜜汁(面向高一)树形dp水题 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct nodeTreeDP { struct node {int nxt, to, w;} E[203]; int n, q, cnt, point[103], apple[103], left[103], ri…

题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1018 Dynamic Programming. 首先要根据input建立树形结构,然后在用DP来寻找最佳结果.dp[i][j]表示node i的子树上最多保存j个分支的最佳结果.代码如下: #include <iostream> #include <math.h> #include <stdio.h> #include <cstdio> #incl…

题目大概说一棵n结点二叉苹果树,n-1个分支,每个分支各有苹果,1是根,要删掉若干个分支,保留q个分支,问最多能保留几个苹果. 挺简单的树形DP,因为是二叉树,都不需要树上背包什么的. dp[u][k]表示以u结点为根的子树保留k个分支最多能有的苹果数 转移就是左子树若干个,右子树若干个转移.. #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define MAXN…

组队赛的时候的一道题,那个时候想了一下感觉dp不怎么好写呀,现在写了出来,交上去过了,但是我觉得我还是应该WA的呀,因为总感觉dp的不对. #pragma warning(disable:4996) #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<string> #include<vector> #define max…

链接 简单树形DP #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> using namespace std; #define INF 0xfffffff ][]; ][]; ][]; int dfs(int pre,int u,int s) { int i; ) return dp[u][s]; ,cc[];…

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> using namespace std; ; int dp[maxn][maxn]; //dp[i][j]表示以i为根,保留j个点的最大权值. int N,Q; int G[maxn][maxn]; int num[maxn]; //以i为根的树的节点个数. //这里…