Hduoj1024【DP】

/*Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18902    Accepted Submission(s): 6217


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.
 
Author
JGShining（极光炫影）

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*/ 
#include<stdio.h>
#include<string.h>
const int MAX = 1000005;

int num[MAX], pre[MAX];

int max(int x, int y)
{
	return x>y?x:y;
}
int DP(int n, int m)
{
	for(int i = 1; i <= m; ++i)//段数 
	{
		int temp = 0;
		for(int j = 1; j <= i; ++j)
		temp += num[j];//即dp[i][j] j == i 
		pre[n] = temp;//临时存放为下一个段数做准备 
		for(int k = i+1; k <= n; ++k)//算出dp[i][k]的最大值 
		{
			temp = max(pre[k-1], temp) + num[k];//dp[i][k] = max(dp[i][t], dp[i][k-1] )+num[k]
			pre[k-1] = pre[n];//保存dp[i][t]的最大值，为下个段数做准备 
			pre[n] = max(pre[n], temp);//更新pre为下一轮i做准备 
		} 
	}
	return pre[n];
}
int main()
{
	int n, m, i;
	while(scanf("%d%d", &m, &n) != EOF)
	{
		for(i = 1; i <= n; ++i)
		scanf("%d", &num[i]);
		memset(pre, 0, sizeof(pre));
		printf("%d\n", DP(n, m));
	}
	return 0;
}

题意：给出n个数，并且给出m条段数，要求每几个连续的数作为一个段，并且段与段之间没有交集，求着n个数中怎样的m段能获得最大的段和。

思路：dp【i】【j】代表前j个数分成i段所能获得的最大和并且最后一段必定包含num【j】，dp【i】【j】 = max（dp【i-1】【t】，dp【i】【j-1】）+num【j】， （i-1<=t<=j-1），dp【i-1】【t】表示当分成i-1段时所能获得的最大值加上当前的num【j】独立段，或者是前j-1个数分成了i段，将num【j】分入最后一段。

这个是整体的思路，但是由于数据量比较大所以这个方法这里要超时，所以需要优化，即将dp【i-1】【t】这个优化可以一边求dp【i】【j】一边求出下一轮i+1所需要的dp【i】【t】。具体求解看这个链接：http://www.cnblogs.com/dongsheng/archive/2013/05/28/3104629.html

难点：如何设计dp方程以及优化是本题的难点。