Description There is a worker who may lack the motivation to perform at his peak level of efficiency because he is lazy. He wants to minimize the amount of work he does (he is Lazy, but he is subject to a constraint that he must be busy when there is…

题目链接:http://poj.org/problem?id=3280 题目大意:给你一个字符串,你可以删除或者增加任意字符,对应有相应的花费,让你通过这些操作使得字符串变为回文串,求最小花费.解题思路:比较简单的区间DP,令dp[i][j]表示使[i,j]回文的最小花费.则得到状态转移方程: dp[i][j]=min(dp[i][j],min(add[str[i]-'a'],del[str[i]-'a'])+dp[i+1][j]); dp[i][j]=min(dp[i][j],min(add[…

题目链接:http://poj.org/problem?id=3186 题目大意:给出的一系列的数字,可以看成一个双向队列,每次只能从队首或者队尾出队,第n个出队就拿这个数乘以n,最后将和加起来,求最大和. 解题思路:有两种写法: ①这是我一开始想的,从外推到内,设立数组dp[i][j]表示剩下i~j时的最优解,则有状态转移方程: dp[i][j]=dp[i][j]=max(dp[i-1][j]+a[i-1]*(n-(j-i+1)),dp[i][j+1]+a[j+1]*(n-(j+1-i)))…

http://poj.org/problem?id=2955 题意:给出一串字符,求括号匹配的数最多是多少. 思路:区间DP. 对于每个枚举的区间边界,如果两边可以配对成括号,那么dp[i][j] = dp[i+1][j-1] + 2,表示由上一个状态加上当前的贡献. 然后和普通的区间合并一样去更新. #include <cstring> #include <cstdio> #include <iostream> #include <string> usin…

题意:中文题面 思路:不知道直接暴力枚举所有情况行不行... 我们可以把答案转化为 所以答案就是求xi2的最小值,那么我们可以直接用区间DP来写.设dp[x1][y1][x2][y2][k]为x1 y1 到 x2 y2 区间分割为k份的最下平方和,显然k = 1是就是区间和的平方. 写了6层for,写出来自己都不信... 交C++才过... 代码: #include<cmath> #include<stack> #include<cstdio> #include<…

Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10010   Accepted: 6188 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one…

题意: 给出一个序列,共n个正整数,要求将区间[2,n-1]全部删去,只剩下a[1]和a[n],也就是一共需要删除n-2个数字,但是每次只能删除一个数字,且会获得该数字与其旁边两个数字的积的分数,问最少可以获得多少分数? 思路: 类似于矩阵连乘的问题,用区间DP来做. 假设已知区间[i,k-1]和[k+1,j]各自完成删除所获得的最少分数,那么a[k]是区间a[i,j]内唯一剩下的一个数,那么删除该数字就会获得a[k]*a[i-1]*a[i+1]的分数了.在枚举k的时候要保证[i,j]的任一子区…

Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [S] are both regular sequences. 3. If A and B are regular sequences, then AB is a regular…

题意:最多有多少括号匹配 思路:区间dp,模板dp,区间合并. 对于a[j]来说: 刚開始的时候,转移方程为dp[i][j]=max(dp[i][j-1],dp[i][k-1]+dp[k][j-1]+2), a[k]与a[j] 匹配,结果一组数据出错 ([]]) 检查的时候发现dp[2][3]==2,对,dp[2][4]=4,错了,简单模拟了一下发现,dp[2][4]=dp[2][1]+dp[2][3]+2==4,错了 此时2与4已经匹配,2与3已经无法再匹配. 故转移方程改为dp[i][j]=…

Blocks Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 4250   Accepted: 1704 Description Some of you may have played a game called 'Blocks'. There are n blocks in a row, each box has a color. Here is an example: Gold, Silver, Silver, Sil…