hdu1024——Max Sum Plus Plus（DP）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39886 Accepted Submission(s): 14338

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 $S_1, S_2, S_3, S_4 … S_x, … S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767)$. We define a function $sum(i, j) = S_i + … + S_j (1 ≤ i ≤ j ≤ n)$.

Now given an integer $m (m > 0)$, your task is to find m pairs of i and j which make $sum(i_1, j_1) + sum(i_2, j_2) + sum(i_3, j_3) + … + sum(i_m, j_m)$ maximal ($i_x ≤ i_y ≤ j_x$ or $i_x≤ j_y ≤ j_x$ 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 $sum(i_x, j_x)(1 ≤ x ≤ m)$ instead.

Input

Each test case will begin with two integers $m$ and $n$, followed by $n$ integers $S_1, S_2, S_3 … S_n$.
Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 2	1 3 1 2 3 2 6 -1 4 -2 3 -2 3

Sample Output

1
2

6
8

题意

将一个长度为$n$的数组分成不相交的$m$段，求这$m$段的和的最大值

思路

状态：$dp[i][j]$表示在前$j$个数中取出$i$段的最大和

状态转移方程：$dp[i][j]=max(dp[i-1][k],dp[i][j-1])+num[j] (i-1\leq k \leq j-1)$

由于$m$范围未知，$n\leq 10^6$，所以二维的dp方程无论是在时间上还是在空间上都是不允许的。

那么我们就需要对这个方程进行优化：

不难发现当前状态只与两个状态有关：

第$j$个数和前$j-1$个数在一段里
第$j$个数和前$j-1$个数不在一段里。

根据这一点，我们把状态降成一维的数组，$dp[j]$表示前$j$个数分$i$段时的最大和，然后用$sum[j-1]$来表示状态一的前$j-1$个数在前$i-1$段的最大和，$dp[j-1]$表示状态二的前$j-1$个数在前$i$段的最大和。

当前状态的转移方程为：$dp[j]=max(dp[j-1],sum[j-1])+num[j]$，持续更新dp与sum数组的值

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#include <time.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
#define bug cout<<"-------------"<<endl
#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"
const double E=exp(1);
const int maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
int a[maxn];
int dp[maxn];
int sum[maxn];
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	#ifndef ONLINE_JUDGE
	    freopen("in.txt", "r", stdin);
	    freopen("out.txt", "w", stdout);
	    double _begin_time = clock();
	#endif
	int k,n;
	while(cin>>k>>n)
	{
		int res;
		for(int i=1;i<=n;i++)
			cin>>a[i];
		ms(dp,0);
		ms(sum,0);
		for(int i=1;i<=k;i++)
		{
			res=-INF;
			for(int j=i;j<=n;j++)
			{
				dp[j]=max(sum[j-1],dp[j-1])+a[j];
				sum[j-1]=res;
				res=max(res,dp[j]);
			}
		}
		cout<<res<<endl;
	}
	#ifndef ONLINE_JUDGE
	    long _end_time = clock();
	    printf("time = %lf ms.", _end_time - _begin_time);
	#endif
	return 0;
}