这题的意思就是取m个连续的区间,使它们的和最大,下面就是建立状态转移方程 dp[i][j]表示已经有 i 个区间,最后一个区间的末尾是a[j] 那么dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][1..j-1])+a[j]) 看数据范围,1e6 肯定开不下数组,观察发现,dp[i][j]仅和dp[i][j-1]和dp[i-1][1..j-1]中最大值有关,即只和dp[i-1]有关 所以开滚动数组求解   复杂度可以通过开数组mmax[j]表示dp[i-1][1.…

Problem Description 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 sequ…

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,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 题目大意:有多组输入,每组一行整数,开头两个数字m,n,接着有n个数字.要求在这n个数字上,m块数字的最大和.比如2 6 -1 4 -2 3 -2 3,就是(4 -2 3)和(3)这两块最大和为8. 解题思路:当成有m层,我们可以设置两个数组dp,mpre.dp[j]记录当前这一层包含a[j]时的最大值(包含a[j]),mpre[j]个记录上一层到第j-1个位置时的最大和(不一定包含a[j])…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1024 题意: 给定序列,给定m,求m个子段的最大和. 分析: 设dp[i][j]为以第j个元素结尾的i个子段的和. 对于每个元素有和前一个元素并在一起构成一个子段,和单独开启一个子段两种可能,状态转移方程 dp[i][j] = max(dp[i][j - 1], dp[i - 1][k]) + a[j] (k >= i - 1 && k <= j - 1) 时间复杂度O(m∗n2…

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面 方程: a[j]直接接在前面…

HDU 1024 Max Sum Plus Plus (动态规划) Description 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…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35988    Accepted Submission(s): 12807 Problem Description Now I think you ha…

Max Sum Plus Plus Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1024 Description 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 mor…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34541    Accepted Submission(s): 12341 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

A - Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a bra…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29942    Accepted Submission(s): 10516 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25639    Accepted Submission(s): 8884 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

Problem Description 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 sequ…

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 207989 Accepted Submission(s): 48681 Problem Description Given a sequence a[1],a[2],a[3]--a[n], your job is to calculate the max sum of a su…

Max Sum Plus Plus Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22262    Accepted Submission(s): 7484   Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. T…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17336    Accepted Submission(s): 5701 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37418    Accepted Submission(s): 13363 Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem.…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44371    Accepted Submission(s): 16084 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18653    Accepted Submission(s): 6129 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

Problem Description 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 sequ…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1024 题意:给定一个数组,求其分成m个不相交子段和的最大值. 这题有点问题其实m挺小的但题目并没有给出. dp[i][j]表示取第i 位的数共取了j段然后转移方程显然为 dp[i][j]=max(dp[i - 1][j]+a[j] , max(dp[j - 1][j - 1]~dp[i - 1][j - 1]))(大致意思是取第i位要么i-1位取了j个那么a[j]刚好能与i-1拼成一段,或者j -…

传送门 题目大意: 给一个序列,要求将序列分成m段,从左至右每一段分别长l1,l2,...lm,求最大的和是多少. 题目分析: 和最大m段子段和相似,先枚举\(i \in [1,m]\),然后$j \in [num[m], n] $,dp转移为: \[dp[j][i] = max(dp[j - 1][i], dp[j - num[i]][i - 1] + sum[j] - sum[i - num[i]])\] code: #include<iostream> #include<cstdi…

如何确保每个段至少一个数是关键(尤其注意负数情况) #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<string> #include<vector> #include<stack> #include<queue…

今天看了一上午dp.看不太懂啊.dp确实不简单.今天開始学习dp,搜了杭电的dp46道,慢慢来吧.白书上的写的 又不太具体,先写几道题目再说. .. 题目连接:id=516&page=1">点击打开链接 思路:就是当当前的和是小于0的时候就又一次计数.大于或者等于0的时候都相加... id=516&page=1">代码: /* Name: Copyright: Author: Date: 08/08/15 08:41 Description: */ #inc…

感觉有点奇怪的是这题明明是n^2的复杂度,n=1e6竟然能过= =.应该是数据水了. dp[i][j]表示前j个数,分成i段,且最后一段的最后一个为a[j]的答案.那么转移式是:dp[i][j] = max(dp[i][j-1], max{dp[i-1][t]}) + a[j],(i-1<=t<=j-1,j-1>=i).前者表示在第i段的最后一个加上a[j],后者表示a[j]另起一段.这个dp显然是可以滚动数组的,那么空间是可以接受的.然后后者可以使用一个pre数组来记录之前的最大值.具…

给定n个数字,求其中m段的最大值(段与段之间不用连续,但是一段中要连续) 例如:2 5 1 -2 2 3 -1五个数字中选2个,选择1和2 3这两段. dp[i][j]从前j个数字中选择i段,然后根据第j个数字是否独立成一段,可以写出 状态转移方程:dp[i][j]=max(dp[i][j-1]+num[j],max(dp[i-1][k])+num[j]) 这里的max(dp[i-1][k])代表的拥有i-1段时的最大值,然后再加上num[j]独立成的一段. 但是题目中没有给出m的取值范围,有可…

转载请注明出处,谢谢:http://www.cnblogs.com/KirisameMarisa/p/4302208.html   ---by 墨染之樱花 dp是竞赛中常见的问题,也是我的弱项orz,更要多加练习.看到邝巨巨的dp专题练习第一道是Max Sum Plus Plus,所以我顺便把之前做过的hdu1003 Max Sum拿出来又做了一遍 HDU 1003 Max Sum 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003 题目描述:…

虽然这道题看起来和 HDU 1024  Max Sum Plus Plus 看起来很像,可是感觉这道题比1024要简单一些 前面WA了几次,因为我开始把dp[22][maxn]写成dp[maxn][22]了,Orz 看来数组越界不一定会导致程序崩溃,也有可能返回一个错误的结果 dp[i][j]表示前j个数构成前i段所得到的最大值 状态转移方程: dp[i][j] = max{dp[i][j-1],  dp[i-1][j-len[i]] + sum[j] - sum[j-len[i]]} 分别对应…

http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Problem Description 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 fac…