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Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17336 Accepted Submission(s): 5701


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(im

, 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.

給定一個數組，求M段連續的子段和最大。dp[i][j]表示前i段選擇第j個元素的最優解。

dp[i][j]=max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j

由於k<j，所以我們能夠用兩個一維數組，一個dp[i]記錄當前行狀態。一個d[i]記錄下一行可選的最大值。

用64位會超時。但int能水過。。

</pre><pre name="code" class="cpp">/*
dp[i][j]=max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j
第j個元素選i段區間，對於當前元素a[j],能夠把它接到第i段的第j-1位置構成一段。
或者單獨成一段。

觀察這個方程。對於dp[i][j-1]這一項，它和dp[i][j]同i,就是在一段。
用一維數組就能記錄；
dp[i-1][k]是i-1段j位置前能取的最大值。我們能夠在計算當前第i段時，
把dp[i-1][k]記錄下來，取最大值就好了。
*/

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define ll __int64
const int inf=0x7fffffff;
#define N 1000010
int a[N];
int pre[N];        //即dp[i-1][k]。記錄j位置前可選值
int dp[N];
int main()
{
    int i,j,n,m;
    int tmp,ans;
    while(~scanf("%d%d",&m,&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            dp[i]=0;
            pre[i]=0;
        }
        pre[0]=dp[0]=0;
        ans=-inf;
        for(i=1;i<=m;i++)
        {
            tmp=-inf;
            for(j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j]);
                pre[j-1]=tmp;      //tmp記錄的是當前段j位置的最大值，
                tmp=max(tmp,dp[j]);//而數組pre[]是記錄j-1位置的，所以要先取tmp,再更新tmp
            }
        }
        for(i=m;i<=n;i++)
            ans=max(ans,dp[i]);
        printf("%d\n",ans);
    }
    return 0;
}

hdu 1024 Max Sum Plus Plus (子段和最大問題)