题意:有n种立方体 每种都有无穷多个 要求选一些立方体叠成一根尽量高的柱子  (可以自行选择哪条边为高 )使得每个立方体的底面都严格小于他下方的立方体 为DAG模型 在任何时候 只有顶面的尺寸会影响到后续决策!!!! 可以采用a,b来表示顶面尺寸   不过落实到dp会有一个问题: 因为a,b的值可能会很大  所以用(idx,k)来表示  idx为立方体的编号  k为第几条边作为高 dp[i][j]表示以 第i个立方体  第k条边为高(注意有序化)  作为最底下的立方体的最大高度 所以可以很轻松得…

据说是DAG的dp,可用spfa来做,松弛操作改成变长.注意状态的表示. 影响决策的只有顶部的尺寸,因为尺寸可能很大,所以用立方体的编号和高的编号来表示,然后向尺寸更小的转移就行了. #include<bits/stdc++.h> using namespace std; #define MP make_pair #define fi first #define se second typedef pair<int,int> pii; ; ]; ]; ]; ][]; int mai…

[Solution] 接上一篇,在处理有向无环图的最长链问题的时候,可以在做拓扑排序的同时,一边做DP; 设f[i]表示第i个方块作为最上面的最高值; f[y]=max(f[y],f[x]+h[y]);(x−>y)∈E 这样可以保证,按阶段进行DP,每次在获取f[x]的时候,你可以保证f[x]已经获得了; 最后取max(f[1..n]) [Code] #include <bits/stdc++.h> using namespace std; #define lson l,m,rt<…

传送门 Description Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story: The babylonians had n…

UVA437 The Tower of Babylon 题解 初始时给了 \(n\) 种长方体方块,每种有无限个,对于每一个方块,我们可以选择一面作为底.然后用这些方块尽可能高地堆叠成一个塔,要求只有一个方块的底的两条边严格小于另一个方块的底的两条边,这个方块才能堆在另一个上面. 问题的思考在于每种方块有无限个,如果我们直接利用该条件问题会变得比较复杂.其实仔细考虑方块堆叠的要求,会发现这是一个约束很强的条件. 注意到,方块堆叠的要求描述的对象不只是方块本身,更细地说,它应该描述的是方块摆放方式…

Problem UVA437-The Tower of Babylon Accept: 3648  Submit: 12532Time Limit: 3000 mSec Problem Description Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the ed…

https://odzkskevi.qnssl.com/5e1fdf8cae5d11a8f572bae96d6095c0?v=1507521965 Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contes…

The Tower of Babylon My Tags Cancel - Seperate tags with commas. Source : University of Ulm Internal Contest 1996 Time limit : 5 sec Memory limit : 32 M Submitted : 303, Accepted : 155 Perhaps you have heard of the legend of the Tower of Babylon. Now…

 The Tower of Babylon  Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story: The babylonian…

转自:https://mp.weixin.qq.com/s/oZVj8lxJH6ZqL4sGCXuxMw The Tower of Babylon(巴比伦塔) Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this…

Description   Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story: The babylonians had n t…

The Tower of Babylon Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 437 Appoint description:  System Crawler  (2015-08-29) Description   Perhaps you have heard of the legend of the Tower of Babylon.…

The Tower of Babylon Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2207   Accepted: 1244 Description Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line…

题目:The Tower of Babylon 这是一个DAG 模型,有两种常规解法 1.记忆化搜索, 写函数,去查找上一个符合的值,不断递归 2.递推法 方法一:记忆化搜索 #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; struct node { int x…

 题意  给你n种长方体  每种都有无穷个  当一个长方体的长和宽都小于还有一个时  这个长方体能够放在还有一个上面  要求输出这样累积起来的最大高度 由于每一个长方体都有3种放法  比較不好控制   能够把一个长宽高分成三个长方体  高度是固定的  这样就比較好控制了 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define maxn 105 int x[…

The Tower of Babylon 题意:给你n种石头,长x,宽y,高z,每种石头数目无限,一块石头能放到另一块上的条件是:长和宽严格小于下面的石头.问叠起来的最大高度. /* 有些类似“叠箱子”问题,看起来一种砖有无限多个,其实最多只能用到两次. 说下我的思路吧,一块砖有3个数据,虽然3!＝6,但本质上还是只有3种,把这三种都表示出来,使x<=y:这样就有了3n组数据.因为我不会建图,就把这3n组数据再排列一下,使一块砖只能放到它后面的砖之上,而绝不能放到之前的砖上,即按x为一级y为二级…

The Tower of Babylon Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story: The babylonians…

The Tower of Babylon Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on PKU. Original ID: 224164-bit integer IO format: %lld      Java class name: Main Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many d…

Labyrinth Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1507    Accepted Submission(s): 520 Problem Description 度度熊是一只喜欢探险的熊,一次偶然落进了一个m*n矩阵的迷宫,该迷宫只能从矩阵左上角第一个方格开始走,只有走到右上角的第一个格子才算走出迷宫,每一次只能走一格,…

Break Standard Weight Time Limit: 2 Seconds                                     Memory Limit: 65536 KB                             The balance was the first mass measuring instrument invented. In its traditional form, it consists of a pivoted horizon…

LightOJ 1033  Generating Palindromes(dp) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/A 题目: Description By definition palindrome is a string which is not changed when reversed. "MADAM" is a nice example of palindrome.…

lightOJ 1047   Neighbor House (DP) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/C 题目: Description The people of Mohammadpur have decided to paint each of their houses red, green, or blue. They've also decided that no two n…

C. Boredom time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a ga…

题目链接:http://poj.org/problem?id=2479 题意:求所给数列中元素值和最大的两段子数列之和. 分析:从左往右扫一遍,b[i]表示前i个数的最大子数列之和. 从右往左扫一遍,c[i]表示后i个数的最大子数列之和. ans=max(ans,b[i]+c[i+1])0<i<n dp方程为 sum=max(sum+a[i],a[i]) dp[i]=max(dp[i-1],sum) #include <cstdio> #include <cstring>…

UVA11125 - Arrange Some Marbles(dp) option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2066" target="_blank" style="">题目链接 题目大意:给你n种不同颜色的弹珠.然后给出每种颜色的弹珠的个数,如今要求你将这些弹珠排序,要求同样颜色的部分最多3个.然后同样颜色的弹珠称…

题意:给出按照x坐标排序的n个点,让我们求出从最左端点到最右短点然后再回来,并且经过所有点且只经过一次的最短路径. 分析:这个题目刘汝佳的算法书上也有详解(就在基础dp那一段),具体思路如下:按照题目的描述比较难考虑,不如把这个问题想成两个人,分别从最左端走到最右端并且不走到重复的点所需要的最小路程,我们定义dp[i,j]表示第一个人走到第i个位置,第二个人走到第j个位置所耗费的路程,并且让第一个人的位置大于第二个人的位置,且前i个城市都已经被走过,那这样的话能走的点只有i+1,i+2,……n这…

有形如下图所示的数塔,从顶部出发,在每一结点可以选择向左走或是向右走,一起走到底层,要求找出一条路径,使路径上的值最大. 样例输入: 5 13 11 8 12 7 26 6 14 15 8 12 7 13 24 11 样例输出: 86(13->8->26->15->24) #include <iostream> #include <cstdio> #include <cstring> #define maxn 105 using namespac…

题意:一串括号,每个括号代表一个值,当有相邻括号组成()时,可以交换他们两个并得到他们值的乘积,问你最大能得到多少 思路:DP题,注定想得掉头发. 显然一个左括号( 的最远交换距离由他右边的左括号的最终位置决定,那么我们可以从右边开始做.我们用dp[i][j]表示第i个左括号交换到第j个位置后,他和他后面左括号所能得到的最大值.显然,dp[i][j] = i交换得到的值 + 后面左括号产生的最大值.而后面左括号能产生的最大值显然就是max(dp[i+1][k])其中j <= k <= n. 代…

1. 问题描述 子串应该比较好理解,至于什么是子序列,这里给出一个例子:有两个母串 cnblogs belong 比如序列bo, bg, lg在母串cnblogs与belong中都出现过并且出现顺序与母串保持一致,我们将其称为公共子序列.最长公共子序列(Longest Common Subsequence,LCS),顾名思义,是指在所有的子序列中最长的那一个.子串是要求更严格的一种子序列,要求在母串中连续地出现.在上述例子的中,最长公共子序列为blog(cnblogs,belong),最长公共子…

[POJ 3071] Football(DP) Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4350   Accepted: 2222 Description Consider a single-elimination football tournament involving 2n teams, denoted 1, 2, -, 2n. In each round of the tournament, all tea…