hdu 1024Max Sum Plus Plus 最大M段字段和单调优化DP 从一类单调性问题看算法的优化

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Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11231 Accepted Submission(s): 3691

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 ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ j_xor 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 S₁, S₂, S₃ ... S_n.
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

#include<stdio.h>
#include<string.h>
#define size 1000100
#define INF 1<<30
int dp[2][size];
int num[size];
int k,n;
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int i,j,l,ans;
	while(scanf("%d %d",&k,&n))
	{
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++)
		{
			scanf("%d",&num[i]);
			dp[1][i]=dp[0][i]=0;
		}
		for(i=1;i<=k;i++)
		{
			ans=-INF;
			for(j=i;j<=n;j++)//前j-1个取i段;前j-1个取i-1段(上一状态)
			{                //接着放  ，自己另起一段，此时dp[0][j-1]中存的还是前j-1个数组成i-1的最大值
				dp[1][j]=max(dp[1][j-1],dp[0][j-1])+num[j];//dp[1]就是已经把这个数放了
				dp[0][j-1]=ans;//用过再更新，变成前j-1个数存i段的最大和，不影响下一层，到下一层就是j-2了
				if(ans<dp[1][j])//若这个数不该放，前面dp[1][j]加上了num[j],也不改变ans的值
				{
					ans=dp[1][j];  //ans存的是i段的最优结果，把每次的最大和存在dp[0]中 
				}                  
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}