ACM学习感悟——HDU1204

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
这道题还是有点让我头疼的，主要是状态转移方程有点让人费解：dp[i][j]=MAX(dp[i][j-1] , dp[i-1][k])+data[j]
其中i<k<=j。 表示前j个数分成i段的最大值。而且由于数据规模较大，要使用滚动数组来节省内存。

<span style="font-size:14px;color:#330000;">///////////////////////////////////////////////////////// 
//   HDU  2844     dp                                  //                        //
//  Created by 吴尔立 			                       //
//  Copyright (c) 2015年 吴尔立. All rights reserved.  //
/////////////////////////////////////////////////////////
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>           
#include <algorithm>
#include <cctype>
#include <stack>
#include <queue>
#include <map>
#include <string>
#include <set>
#include <vector>
#define ll long long;
#define INF 1<<31
#define cir(i,a,b)  for (int i=a;i<=b;i++)
#define CIR(j,a,b)  for (int j=a;j>=b;j--)
#define CLR(x) memset(x,0,sizeof(x))
using namespace std;
#define maxn 1000005
int n,m;
int a[maxn],d[maxn],td[maxn];           //td表示前一个状态的最大值； d[j]表示前j个元素分成i段的最大值 

int main()
{
	while (scanf("%d%d",&m,&n)!=EOF)
	{
		CLR(d);
		CLR(td);
		cir(i,1,n) scanf("%d",&a[i]);
		int temp;
		cir(i,1,m)
		{
			temp=-INF;
			cir(j,i,n)
			{
				d[j]=max(d[j-1],td[j-1])+a[j];    //这里有点难以理解，若是d+a[i]，则代表是将ai并上去，否则是独立成一段。</span>

<span style="font-size:14px;color:#330000;">				td[j-1]=temp;    //temp一直表示前一个状态的最大值。
				temp=max(temp,d[j]);
			} 
		}
		printf("%d\n",temp);
	}
}</span>