/* dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) (0<k<j) dp[i][j-1]+a[j]表示的是前j-1分成i组,第j个必须放在前一组里面. max( dp[i-1][k] ) + a[j] )表示的前(0<k<j)分成i-1组,第j个单独分成一组. */ #include <iostream> #include <cstdio> #include <cstring> #inclu…

Max Sum Plus Plus     HDU - 1024 Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutiv…

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutive number sequence S 1, S 2, S 3,…

用二位数组dp[i][j]记录组数为i,前j个数字的最大子段和. 转移方程: dp[i][j],考虑第j个数,第j个数可以并到前面那一组,此时dp[i][j]=dp[i][j-1]+arr[j],第j个数也可以是作为新的一组,那么dp[i][j]=max(dp[i-1][k])(i-1<=k<=j-1)+arr[j].我们只要求出前i-1组最大的字段和,然后加上arr[j]这一新的组就行了. 二维数组(o(n^3))的写法 ;i<=m;i++) for(int j=i;j<=n;j…

HDU 1024  Max Sum Plus Plus // dp[i][j] = max(dp[i][j-1], dp[i-1][t]) + num[j] // pre[j-1] 存放dp[i-1][t] 里的 (1<=t<=j-1)最大值. //dp[j] = max(dp[j-1], pre[j-1]) + num[j]; #include <stdio.h> #include <string.h> #include <iostream> #defin…

原文:php基础篇-二维数组排序 array_multisort 对2维数组或者多维数组排序是常见的问题,在php中我们有个专门的多维数组排序函数,下面简单介绍下: array_multisort(array1,sorting order, sorting type,array2,array3..)是对多个数组或多维数组进行排序的函数. array1 必需.规定输入的数组. sorting order 可选.规定排列顺序.可能的值是 SORT_ASC 和 SORT_DESC. sorting t…

python3--算法基础:二维数组转90度 [0, 1, 2, 3][0, 1, 2, 3][0, 1, 2, 3][0, 1, 2, 3] 二维数组转90度 [0, 0, 0, 0][1, 1, 1, 1][2, 2, 2, 2][3, 3, 3, 3] #!/usr/bin/env python # -*- coding:utf-8 -*- # 算法基础:二维数组转90度 a = [col for col in range(4)] print(a) print("-------------…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1081 To The Max Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8839    Accepted Submission(s): 4281 Problem Description Given a two-dimensional ar…

Max Sum of Max-K-sub-sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5084    Accepted Submission(s): 1842 Problem Description Given a circle sequence A[1],A[2],A[3]......A[n]. Circle s…

传送门:Max Sum Plus Plus 题意:从n个数中选出m段不相交的连续子段,求这个和最大. 分析:经典dp,dp[i][j][0]表示不取第i个数且前i个数分成j段达到的最优值,dp[i][j][1]表示取了第i个数且前i个数分成j段达到的最优值. 那么有: dp[i][j][0]=max(dp[i-1][j][0],dp[i-1][j][1]). dp[i][j][1]=max(dp[i-1][j-1][0]+a[i],max(dp[i-1][j-1][1],dp[i][j][1])…