hdoj 1024 Max Sum Plus Plus 【动态规划】

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21508    Accepted Submission(s): 7202

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

分析：

求子段最大和。

代码：

#include
#include
#include
#define Min  0x80000000  
using namespace std;
const int  N= 1e6+10; 
int pp[N],bb[N],dp[N];
int main()
{
	int n,m,i,j,k,max;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=m;i++)
		scanf("%d",&pp[i]);
		for(i=0;i<=m;i++)
		dp[i]=0,bb[i]=0;
		for(i=1;i<=n;i++)
		 {
		 	max=Min;
		 	for(j=i;j<=m;j++)
		 	{
		 		if(dp[j-1]>bb[j-1])
		 		dp[j]=dp[j-1]+pp[j];
		 		else dp[j]=bb[j-1]+pp[j];
		 		bb[j-1]=max;
		 		if(max