Covered Walkway Problem Description   Your university wants to build a new walkway, and they want at least part of it to be covered. There are certain points which must be covered. It doesn’t matter if other points along the walkway are covered or no…

Covered Walkway Time Limit: 30000/10000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1273    Accepted Submission(s): 491 Problem Description Your university wants to build a new walkway, and they want at least…

题目大意: 一个N个点的序列,要将他们全部覆盖,求总最少费用:费用计算: c+(x-y)2 分析: 斜率优化DP 我们假设k<j<i.如果在j的时候决策要比在k的时候决策好,那么也是就是dp[j]+M+(sum[i]-sum[j])^2<dp[k]+M+(sum[i]-sum[k])^2.(因为是最小花费嘛,所以优就是小于) 两边移项一下,得到:(dp[j]+num[j]^2-(dp[k]+num[k]^2))/(2*(num[j]-num[k]))<sum[i].我们把dp[j]…

斜率优化的模板题 给出n个数以及M,你可以将这些数划分成几个区间,每个区间的值是里面数的和的平方+M,问所有区间值总和最小是多少. 如果不考虑平方,那么我们显然可以使用队列维护单调性,优化DP的线性方法来做,但是该题要求的是区间和的平方,于是要转换单调的计算方法为斜率,也就是凸线. 其他就是最基本的单调DP /** @Date : 2017-09-04 15:39:05 * @FileName: HDU 3507 单调队列 斜率优化 DP.cpp * @Platform: Windows * @…

题目链接:hdu 3507 Print Article 题意: 每个字有一个值,现在让你分成k段打印,每段打印需要消耗的值用那个公式计算,现在让你求最小值 题解: 设dp[i]表示前i个字符需要消耗的最小值,那么有dp[i]=min{dp[k]+(sum[i]-sum[k])2+m)}(k<i). 这样是n2 的做法. 考虑用斜率优化: 设k<j,对于dp[i],从k+1到i为一段比j+1到i为一段更优. 那么有 dp[j]+(sum[i]-sum[j])2+m<=dp[k]+(sum[…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2829 题目大意:有一段铁路有n个站,每个站可以往其他站运送粮草,现在要炸掉m条路使得粮草补给最小,粮草补给的公式是将每个站能收到的粮草的总和. 4----5-----1-----2 粮草总和为4*5 + 4*1 + 4*2 + 5*1 + 5*2 + 1*2 = 49. 4----5       1-----2 粮草总和为4*5 + 1*2 = 22. 4      5-----1------2 粮…

Print Article Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 4810    Accepted Submission(s): 1451 Problem Description Zero has an old printer that doesn't work well sometimes. As it is antique…

Division Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others) Total Submission(s): 3984    Accepted Submission(s): 1527 Problem Description Little D is really interested in the theorem of sets recently. There’s a pro…

题目链接 \(Description\) 给定一个\(n\)个数的序列,最多将序列分为\(m+1\)段,每段的价值是这段中所有数两两相乘的和.求最小总价值. \(Solution\) 写到这突然懒得写了.. 丢个题解走人 /* 朴素O(n^3):f[i][j]表示当前在i分了j段的最小价值 W[i]表示1~i的总价值 S[i]表示1~i的原序列值之和 则有 f[i][j]=min{ f[k][j-1]+W[i]-W[k]-S[i]*(S[i]-S[k]) } (1≤k<i) 这个方程可以用斜率优…

解题思路 首先肯定是考虑如何快速求出一段铁路的价值. \[ \sum_{i=1}^k \sum_{j=1, j\neq i}^kA[i]A[j]=(\sum_{i=1}^kA[i])^2-\sum_{i=1}^kA[i]^2 \] 那么我们要维护如下两个东西,就可以在\(O(1)\)内求出一段铁路的价值了. for( LL i = 1; i <= N; ++i ) Sum[ i ] = Sum[ i - 1 ] + A[ i ]; for( LL i = 1; i <= N; ++i ) Su…