题目链接：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): 21926 Accepted Submission(s): 7342

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 sequence S₁, S₂, S₃, S₄ ... S_x, ... S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767). We define a function sum(i, j) = S_i + ... + S_j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ j_x or i_x ≤ j_y ≤ j_x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i_x, j_x)(1 ≤ x ≤ m) instead. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1 2 3

2 6 -1 4 -2 3 -2 3

Sample Output

Hint

Huge input, scanf and dynamic programming is recommended.

Author

JGShining（极光炫影）

题意：输入m,n，然后输入n个数。求最大连续m段和。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>using namespace std;const int oo = 0x7fffffff;
int a[1000005];
int dp[1000005];
int Max[1000005];int main()
{int m,n;while(~scanf("%d%d", &m, &n)){for(int i=1; i<=n; i++){scanf("%d", &a[i]);Max[i] = 0;dp[i] = 0;}int M;dp[0] = 0;Max[0] = 0;for(int i=1; i<=m; i++){M = -oo;for(int j=i; j<=n; j++){dp[j] = max(dp[j-1]+a[j], Max[j-1]+a[j]);//其中dp[j-1]表示的是以j-1结尾的元素i个子段的数和，Max[j-1]表示的是前j-1个元素中i-1个子段的数和Max[j-1] = M;//更新Max数组，下次循环用到。放在此处是为了实现Max[j-1]+a[j]中a[j]是一个独立的子段，那么此时就应该用的是i-1段M = max(dp[j], M);//更新M
            }}printf("%d\n",M);}return 0;
}

转载于:https://www.cnblogs.com/mengzhong/p/5058186.html

HDU 1024 Max Sum Plus Plus相关推荐

HDU 1024 Max Sum Plus Plus 动态规划
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 题目大意:n个数分成两两不相交的m段,求使这m段和的最大值. 解题思路:比较坑的点:n2 能过: ...
HDU 1024 ~ Max Sum Plus Plus (DP + XJB优化)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 题意:给你 n 个数字,你可以从中选m个子段(子段不能相交).让选出的子段和和最大. 题解:明显 ...
hdu 1024 Max Sum Plus Plus(dp 最大m子段和)
题意是输入m,n. m为你要求的子段个数,n为数据个数. 由于是很早的题型了,但是理解起来还是很是无力. 并于是用了三天来搞懂此类问题.发现网上大多代码无思路整个过程. 就大致讲解一下DP的整个思路. ...
HDU - 1024 Max Sum Plus Plus 最大m段子段和+滚动数组优化
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 ...
HDU 1024 Max Sum Plus Plus
本题的大致意思为给定一个数组,求其分成m个不相交子段和最大值的问题. 读完题就能看出要用DP,但是因为m的范围没有限定,= =加上自己太菜不知道怎么写, 总之先膜拜大佬http://www.cnblo ...
HDU 1244 Max Sum Plus Plus Plus
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][2 ...
HDU.1003 Max Sum
原题 HDU.1003 Max Sum 分类动态规划题意计算从一个序列中最大连续子序列和.对应的起始元素和终止元素的位置. 输入/输出要求与格式样例数的确定最开始一行开始输入样例数每个样 ...
HDUOJ 1024 Max Sum Plus Plus
HDUOJ 1024 Max Sum Plus Plus 题目链接 Problem Description Now I think you have got an AC in Ignatius.L's ...
[Hdoj 1024] Max Sum Plus Plus
链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 Problem Description Now I think you have got an AC ...

HDU 1024 Max Sum Plus Plus

Max Sum Plus Plus

HDU 1024 Max Sum Plus Plus相关推荐

最新文章

热门文章