HDU-1024 Max Sum Plus Plus ( DP )

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1024

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

 

题目大意 在长度为n的序列中找m个不相交的连续子序列使其和最大，输出最大和

状态转移 dp[i][j] = max(dp[i - 1][k] + num[j] ),( 1 <= k < j ),ans = max( dp[i][k] ),( 1 <= k <= n ),dp[i][j]表示取num[j]的情况下前j个数字i分的最大值;但是复杂度为n^3，考虑到求dp[i][j]时所用的dp[i][k]即为前j-1个数i分时的ans，可以用两个数组分别存储，将复杂度优化为n^2。

 

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

int num[1000005];
int dp[1000005];
int mmax[1000005];
int ans;
int m, n;

int main(){
    ios::sync_with_stdio( false );

    while( cin >> m >> n ){
        memset( dp, 0, sizeof( dp ) );
        memset( mmax, 0, sizeof( mmax ) );

        for( int i = 1; i <= n; i++ )
            cin >> num[i];

        for( int i = 1; i <= m; i++ ){
            ans = -0x3f3f3f3f;
            for( int j = i; j <= n; j++ ){
                dp[j] = max( dp[j - 1] + num[j], mmax[j - 1] + num[j] );
                mmax[j - 1] = ans;
                ans = max( ans, dp[j] );
            }
        }

        cout << ans << endl;
    }

    return 0;
}