hdu 1024 Max Sum Plus Plus（最大m字段和）

Max Sum Plus PlusTime Limit:1000MS    Memory Limit:32768KB    64bit
IO Format:%I64d & %I64u

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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.

 

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]+s[i],max(dp[1][j-1]--dp[i-1][j-1])+s[i]);

其中max(dp[1][j-1]--dp[i-1][j-1])为结尾元素在1到i-1中字段数为j-1的最大值

for(j=1; j<=m; j++)
{
mmx=-999999999;
for(i=j; i<=n; i++)
{
dp[i]=max(dp[i-1]+s[i],ans[i-1]+s[i]);//其中dp[i-1]表示的是以i-1结尾的元素j个子段的数和，
ans[i-1]表示的是前i-1个元素中j-1个子段的数和的最大值。
ans[i-1]=mmx;放在此处是为了实现ans[i-1]+a[i]中a[i]是一个独立的子段，那么此时就应该用的是j-1段
if(dp[i]>mmx)
mmx=dp[i];
}

#include <iostream>
#include <string.h>
using namespace std;
const int nmax = 1000005;
int dp[nmax], s[nmax], ans[nmax];
int main()
{
int m, n, i, j;
while(~scanf("%d%d",&m,&n))
{
for(i=1; i<=n; i++)
scanf("%d", &s[i]);
memset(dp, 0, sizeof(dp));
memset(ans, 0, sizeof(ans));
int mmx;
for(j=1; j<=m; j++)
{
mmx=-999999999;
for(i=j; i<=n; i++)
{
dp[i]=max(dp[i-1]+s[i],ans[i-1]+s[i]);
ans[i-1]=mmx;
if(dp[i]>mmx)
mmx=dp[i];
}

}
printf("%d\n",mmx);
}

}

}