Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (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, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.

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.

题目大意：给N个数字，然后取X段的区间，要求这些区间的和最大。注意：区间不能重合。

题解：用f[i][j]表示前i个数第j个不相交的子段的最大值，则f[i][j] = max(f[i]][j],f[i][j-1]+a[i])。但是数组是1e6，会炸。所以用滚动数组，以j为最外层循环，这样f[i][j-1]可以在上一层处理出来。

附上代码

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;const int INF = 1e9+7;const int maxn = 1000050;int a[maxn],dp[maxn],Max[maxn];//dp记录f[i][j-1],Max记录前i个f[i][j-1]最大值。int main(){ int m,n,i,j,Max_sum; while(scanf("%d%d",&m,&n)!=EOF) { for(i=1;i<=n;i++) { scanf("%d",&a[i]); dp[i] = Max[i] = 0; } for(i=1;i<=m;i++) { Max_sum = -INF; for(j=i;j<=n;j++) { dp[j] = a[j] + max(dp[j-1],Max[j-1]); Max[j-1] = Max_sum; Max_sum = max(Max_sum,dp[j]); } } printf("%d\n",Max_sum); } return 0;}