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 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

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;
}