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 DescriptionNow
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. ^_^ InputEach 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. OutputOutput the maximal summation described above in one line. Sample Input1 3 1 2 32 6 -1 4 -2 3 -2 3 Sample Output6
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 = ;
const int inf = 1e9+;int dp[][N][] , 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 =  ; i <= n ; ++i ) {
            scanf("%d",&x[i]);
        }
        int v =  ;
        dp[v][][] =  ;
        dp[v][][] = x[] ;
        for( int i =  ; i < n ; ++i ) {
            for( int j =  ; j <= i +  && j <= m ; j++ ) {
                dp[v^][j][] = dp[v^][j][] = -inf ;
            }
            for( int j = min( m , i ) ; j >=  ; --j ) {
                if( j != i ) {
                    dp[v^][j+][] = MAX( dp[v][j][] + x[i+] , dp[v^][j+][] );
                    dp[v^][j][] = MAX( dp[v][j][] , dp[v^][j][]);
                }
                if( j !=  ) {
                    dp[v^][j][] =  MAX ( dp[v^][j][] , dp[v][j][] + x[i+] ) ;
                    dp[v^][j+][] = MAX ( dp[v^][j+][] , dp[v][j][] + x[i+] ) ;
                    dp[v^][j][] = MAX ( dp[v^][j][] , dp[v][j][] ) ;
                }
            }
            v ^=  ;
        }
        int ans = -inf ;
        if( m < n ) ans = MAX( ans , dp[v][m][] );
        if( m >  ) ans = MAX( ans , dp[v][m][] );
        printf("%d\n",ans);
    }
    return  ;
}