HDU1024——DP——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.

 

 

Author

JGShining（极光炫影）

 

 

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大意：主要是降低复杂度，问给你m个数要可以分成n块连续的数，问最大和是多少

定义dp[i][j]表示当前这个数为i，已经分了j块

可以转来的方向  1.块数不动多一个数   2.块数动了多一个数

动态转移方程  dp[i][j] = max(dp[i-1][j]+a[i],max(dp[1][j-1].....dp[i-1][j-1])+a[i]))

bing神代码好流弊。。定义一个数组记录前面一个块的前j-1中最大的值

那么最后答案就是这个max2（分成n组时1到j中最大的值

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

int dp[maxn];

int a[maxn],max1[maxn];

int main()

{

    int n,m;

    int max2;

    while(~scanf("%d%d",&n,&m)){

        for(int i = 1; i <= m ;i++)

            scanf("%d",&a[i]);

        memset(dp,0,sizeof(dp));

        memset(max1,0,sizeof(max1));

        for(int i = 1; i <= n; i++){

             max2 = -inf;

            for(int j = i; j <= m; j++){

                   dp[j] = max(dp[j-1]+a[j],max1[j-1]+a[j]);

                   max1[j-1] = max2;

                   max2 = max(max2,dp[j]);

            }

        }

        printf("%d\n",max2);

    }

    return 0;

}