hdu Max Sum Plus Plus(最大m段子段和)

1、http://acm.hdu.edu.cn/showproblem.php?pid=1024

2、题目大意：

已知有n个数，求m段不相交的子段权值之和最大，

状态转移方程：dp[i][j]表示以i为结尾元素的j个子段的数和

dp[i][j]=max(dp[i-1][j]+a[i],dp[i-k][j-1]+a[i]);其中(j-1<=k<=n-m+1)

此题实现这种思想：

for(i=1;i<=m;i++)
{
maxx=(-1)*(N*90);//初始化
for(j=i;j<=n;j++)
{
now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);//其中now[j-1]表示的是以j-1结尾的元素i个子段的数和，pre[j-1]表示的是前j-1个元素中i-1个子段的数和
pre[j-1]=maxx;//放在此处是为了实现pre[j-1]+a[j]中a[j]是一个独立的子段，那么此时就应该用的是i-1段
if(now[j]>maxx)
{
maxx=now[j];
}
}
}

3、题目：

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11549 Accepted Submission(s): 3807

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_xis 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

4、代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
#defineN1000
inta[N+5];
intnow[N+5];
intpre[N+5];
usingnamespacestd;
intmain()
{
intn,m,maxx,i,j;
while(scanf("%d%d",&m,&n)!=EOF)
{
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
memset(now,0,sizeof(now));
memset(pre,0,sizeof(pre));
for(i=1;i<=m;i++)
{
maxx=(-1)*(N*90);
for(j=i;j<=n;j++)
{
now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);
pre[j-1]=maxx;
if(now[j]>maxx)
{
maxx=now[j];
}
}
}
printf("%d\n",maxx);
}
return0;
}