题目链接: Poj 3666 Making the Grade 题目描述: 给出一组数,每个数代表当前位置的地面高度,问把路径修成非递增或者非递减,需要花费的最小代价? 解题思路: 对于修好的路径的每个位置的高度肯定都是以前存在的高度,修好路后不会出现以前没有出现过得高度 dp[i][j]代表位置i的地面高度为第j个高度,然后我们可以把以前的路修好后变成非递减路径,或者把以前的路径首尾颠倒,然后修成非递减路径.状态转移方程为:dp[i][j] = min(dp[i-1][k]) + a[i] -…

Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and valleys. FJ…

Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and valleys. FJ…

Making the Grade Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7068   Accepted: 3265 Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up…

题意:输入N, 然后输入N个数,求最小的改动这些数使之成非严格递增即可,要是非严格递减,反过来再求一下就可以了. 析:并不会做,知道是DP,但就是不会,菜....d[i][j]表示前 i 个数中,最大的是 j,那么转移方程为,d[i][j] = abs(j-w[i])+min(d[i-1][k]);(k<=j). 用滚动数组更加快捷,空间复杂度也低. 代码如下: #include <cstdio> #include <string> #include <cstdlib&…

题意:农夫约翰想修一条尽量平缓的路,路的每一段海拔是A[i],修理后是B[i],花费|A[i] – B[i]|,求最小花费.(数据有问题,代码只是单调递增的情况) #include <stdio.h> #include <algorithm> #include <cstring> #include <cstdlib> #include <cmath> #include <memory> #include <iostream>…

读题堪忧啊,敲完了才发现理解错了..理解题必须看样例啊!! 题目链接: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=110495#problem/S 题意: 给定序列,做出最少的改变,使得新的序列单调非增或者或单调非减. 分析: 先考虑单调非增. 如果后一个元素比前一个小,那么最少改变的情况就是让他和前一个元素相等.如果比前一个元素大或者相等,则不需做出改变. 仔细想想就可以发现其实最后的序列就是由原始数组的元素组成. 那么我们先对…

传送门: http://poj.org/problem?id=3666 Making the Grade Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9468   Accepted: 4406 Description A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would lik…

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)))…