Max Sum Plus Plus HDU - 1024(DP)

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.
제목: n개의 숫자, m 자단과 최대 값.
사고방식: 최대 하위 세그먼트와 업그레이드 버전.최대 자단과 방정식은 dp[i]=max(dp[i-1], 0)+a[i]이다.dp[i]는 끝 숫자가 a[i]인 최대 자단과이 문제는 여러 개의 자단이기 때문에 1차원 상태를 더하면 dp[i][j]이고 i단이 있고 끝 숫자가 a[j]인 결과이다.dp[i][j] = max(dp[i][j - 1],max{dp[i-1][k]}) + a[i]. i -1 ≤ k ≤ j - 1. 1차원 그룹을 생략하고, 하나의 그룹pre로 최대 값 앞 j-1개의 최대 값을 기록합니다.

#include 
#include 
#include 

using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 7;
int dp[maxn],pre[maxn];
int a[maxn];

int main()
{
    int m,n;
    while(~scanf("%d%d",&m,&n))
    {
        memset(dp,0,sizeof(dp));memset(pre,0,sizeof(pre));
        for(int i = 1;i <= n;i++)scanf("%d",&a[i]);
        int ans = -INF;
        for(int i = 1;i <= m;i++)
        {
            ans = -INF;
            for(int j = i;j <= n;j++)
            {
                dp[j] = max(dp[j - 1],pre[j - 1]) + a[j];
                pre[j - 1] = ans;
                ans = max(ans,dp[j]);
            }
        }
        printf("%d
",ans);
    }
    return 0;
}