HDU 1024.Max Sum Plus Plus【动态规划】【11月14】

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 faced with a more difficult problem.

Given a consecutive number sequence S ₁, S ₂, S ₃, S ₄ ... 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(i ₁, j ₁) + sum(i ₂, j ₂) + sum(i ₃, j ₃) + ... + 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 S ₁, S ₂, S ₃ ... 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

   
   
   
   

    
    
    
    6
8


    
    
    
    
 
     
 
      Hint 
     
 Huge input, scanf and dynamic programming is recommended. 
    
    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   

m个不交叉子序列和值最大。

这个我可能解释不清楚，附上各种解题思路：

http://blog.csdn.net/juststeps/article/details/8687369
http://www.cnblogs.com/kuangbin/archive/2011/08/04/2127085.html
http://blog.sina.com.cn/s/blog_677a3eb30100jxqa.html
http://www.xuebuyuan.com/1410276.html
http://blog.csdn.net/acm_davidcn/article/details/5887401
http://www.cnblogs.com/wally/archive/2013/03/13/2957061.html
http://blog.csdn.net/xymscau/article/details/6857311
http://m.blog.csdn.net/blog/kay_zhyu/8727993
http://blog.csdn.net/zhanglei0107/article/details/7423419

// 状态:dp[i][j]表示以j结尾的i个子段和的最优解
// 转移:dp[i][j] = max{dp[i][j-1],max{dp[i-1][t]}}+a[j];
// 第j个数字要么作为新的起点加入队列,要么直接加到前一段里
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <algorithm>
using namespace std;
const int MAXN = 1000000 + 10;
int s[MAXN], DP[MAXN], pre[MAXN], n, m;
int main()
{
    while(scanf("%d%d", &m, &n) != EOF)
    {
        memset(DP, 0, sizeof(DP));
        memset(pre, 0, sizeof(pre));
        for(int i = 1; i <= n; i ++)    scanf("%d", &s[i]);
        int ans;
        for(int i = 1; i <= m; i ++)
        {
            ans = -(1 << 31);
            for(int j = i;j <= n; j ++)
            {
                DP[j] = max(DP[j - 1] + s[j], pre[j - 1] + s[j]);
                // 保存局部最优解
                pre[j - 1] = ans;
                ans = max(ans, DP[j]);
            }
        }
        cout << ans << endl;
    }
    return 0;
}