HDU 1024 Max Sum Plus Plus(最大m子段和)

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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): 30715    Accepted Submission(s): 10845

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 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.

裸的最大m子段和问题，题目最后提示cin，cout可能会超时，还好889ms，差了好几个15ms。

如果有对最大m子段和不了解的可以戳这里：点击打开链接 点击打开链接

最大m子段和，巧妙地将二维dp优化为一维，避免数组太大开不出来的问题。此题为模板题，具体解释在代码中，准备拿最大环形m子段和是谁手，下篇博客见。

代码实现：

#include
#include
#include
#include
#include
#include
#define ll long long
#define mset(a,x) memset(a,x,sizeof(a))

using namespace std;
const double PI=acos(-1);
const int inf=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int mod=1e9+7;

int dp[maxn],map[maxn],mmax[maxn];
//dp[i]代表当前选取i个数中，从1到i的和是多少，mmax储存前一行的数值 

int main()
{
	int n,m,i,j,k;
	while(cin>>m>>n)
	{
		for(i=1;i<=n;i++)
		cin>>map[i];
		
		mset(dp,0);
		mset(mmax,0);
		int maxx;
		for(i=1;i<=m;i++)   //枚举m次选择 
		{
			maxx=-inf;      //初始为最小 
			for(j=i;j<=n;j++)   //枚举每一次选择 
			{
				dp[j]=max(dp[j-1]+map[j],mmax[j-1]+map[j]);  //当前值取前j-1个数的和跟前一行的和的最大值 
				mmax[j-1]=maxx;  //储存前一行 
				maxx=max(dp[j],maxx);  //储存最大值 
			}
		}
		cout<