HDU 1024 Max Sum Plus Plus(DP+预处理优化）

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

题意：给一个含有N个数字的序列，找出M个不相交的子段，使得子段和最大。

思路：设dp[i][j]为前i个数中选出j个子段能得到的最大和，且第j段以a[i]结尾。

若第j段仅由a[i]组成，则dp[i][j]=max(dp[k][j-1])+a[j]。(j-1 <= k <= i-1)

若第j段仅由a[i]与a[i-1]等组成，则dp[i][j]=dp[i-1][j]+a[i]。

首先M未知，二维数组存不了。但是我们可以发现，j状态总是从j-1状态推来的，所以利用滚动数组，可以将dp变为一维数组。

此外，如果用三重循环一定会超时，所以我们需要预处理出max(dp[k][j-1])的部分，设数组imax[i-1]表示max(dp[k][j-1])，j-1 <= k <= i-1。

由于如果我们dp[i]更新后立即更新imax[i]的话，此时imax[i]则表示max(dp[k][j])，当我们枚举到i+1时，就用的是max(dp[k][j])+a[j]，而不是max(dp[k][j-1])+a[j]。这样是错误的。所以我们要先记录下imax[i]的值，当枚举当i+1时再更新imax[i]。

#include <iostream>#include <stdio.h>#include <cmath>#include <algorithm>#include <iomanip>#include <cstdlib>#include <string.h>#include <vector>#include <queue>#include <stack>#include <ctype.h>using namespace std;int n,m;int a[1000005];int dp[1000005];int imax[1000005];int main(){    while(scanf("%d%d",&m,&n)!=EOF)    {        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        memset(imax,0,sizeof(imax));        memset(dp,0,sizeof(dp));        for(int j=1;j<=m;j++)        {            int mmax=-0x3f3f3f3f;            for(int i=j;i<=n;i++)            {                dp[i]=max(dp[i-1]+a[i],imax[i-1]+a[i]);                imax[i-1]=mmax;                mmax=max(mmax,dp[i]);            }        }        int ans=-0x3f3f3f3f;        for(int i=m;i<=n;i++)        {            ans=max(dp[i],ans);        }        cout<<ans<<endl;    }    return 0;}