# Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6725    Accepted Submission(s): 2251

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 Im 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.

Author
JGShining（极光炫影）

Num为给定数组，n为数组中的元素总数，Status[i][j]表示前i个数在选取第i个数的前提下分成j段的最大值，其中1<=j<=i<=n && j<=m，状态转移方程为：

Status[i][j]=Max(Status[i-1][j]+Num[i]Max(Status[0][j-1]~Status[i-1][j-1])+Num[i])

max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个

/*状态dp[i][j]有前j个数，组成i组的和的最大值。决策： 第j个数，是在第包含在第i组里面，还是自己独立成组。方程 dp[i][j]=Max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j空间复杂度，m未知，n<=1000000，  继续滚动数组。 时间复杂度 n^3. n<=1000000.  显然会超时，继续优化。max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个的最大值 用数组保存下来  下次计算的时候可以用，这样时间复杂度为 n^2.*/#include<stdio.h>#include<algorithm>#include<iostream>using namespace std;#define MAXN 1000000#define INF 0x7fffffffint dp[MAXN+10];int mmax[MAXN+10];int a[MAXN+10];int main(){    int n,m;    int i,j,mmmax;    while(scanf("%d%d",&m,&n)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            mmax[i]=0;            dp[i]=0;        }        dp[0]=0;        mmax[0]=0;            for(i=1;i<=m;i++)        {                mmmax=-INF;                for(j=i;j<=n;j++)                {                    dp[j]=max(dp[j-1]+a[j],mmax[j-1]+a[j]);                    mmax[j-1]=mmmax;                    mmmax=max(mmmax,dp[j]);                }            }          printf("%d\n",mmmax);                }     return 0;   }    `

posted on 2011-08-04 11:10 kuangbin 阅读(...) 评论(...) 编辑 收藏

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