HDU 1024 Max Sum Plus Plus (递推)

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18653    Accepted Submission(s): 6129

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.

 

 

dp[i][j][0] ... 表示前i个数分成j个组，不选第i个数的最大得分

dp[i][j][1] ... 表示前i个数分成j个组，选第i个数的最大得分

因为状态i只跟状态i-1, 所以可以用滚动数组来减空间

取最要自己写 。 否则卡常数会超时

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std ;
const int N = 100010;
const int inf = 1e9+7;

int dp[2][N][2] , n , m , x[N] ;
inline int MAX( int a , int b ) {
    if( a > b ) return a ;
    else return b ;
}
int main() {
//    freopen("in.txt","r",stdin);
    while( ~scanf("%d%d",&m,&n) ) {
        for( int i = 1 ; i <= n ; ++i ) {
            scanf("%d",&x[i]);
        }
        int v = 0 ;
        dp[v][0][0] = 0 ;
        dp[v][1][1] = x[1] ;
        for( int i = 1 ; i < n ; ++i ) {
            for( int j = 0 ; j <= i + 1 && j <= m ; j++ ) {
                dp[v^1][j][0] = dp[v^1][j][1] = -inf ;
            }
            for( int j = min( m , i ) ; j >= 0 ; --j ) {
                if( j != i ) {
                    dp[v^1][j+1][1] = MAX( dp[v][j][0] + x[i+1] , dp[v^1][j+1][1] );
                    dp[v^1][j][0] = MAX( dp[v][j][0] , dp[v^1][j][0]);
                }
                if( j != 0 ) {
                    dp[v^1][j][1] =  MAX ( dp[v^1][j][1] , dp[v][j][1] + x[i+1] ) ;
                    dp[v^1][j+1][1] = MAX ( dp[v^1][j+1][1] , dp[v][j][1] + x[i+1] ) ;
                    dp[v^1][j][0] = MAX ( dp[v^1][j][0] , dp[v][j][1] ) ;
                }
            }
            v ^= 1 ;
        }
        int ans = -inf ;
        if( m < n ) ans = MAX( ans , dp[v][m][0] );
        if( m > 0 ) ans = MAX( ans , dp[v][m][1] );
        printf("%d\n",ans);
    }
    return 0 ;
}