Max Sum Plus Plus
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1024
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 ≤ iy ≤ 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, S2, 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：给出m,n,和长度为n的数列，要求出其中m个子段和的最大值
思路：
dp [ i ] [ j ] 为前 j 个数分成 i 组的最大和
状态转移方程：dp[ i ] [ j ]=max (dp [ i ] [ j-1 ] + a[ j ] , dp [ i-1 ] [ j-1 ] + a[ j ]) ;
dp [ i ] [ j-1 ] + a[ j ] 表示前 j-1个数分成 i 组的最大和加 a [ j ] ，就是把第j个数放在前一组里
dp [ i-1 ] [ j-1 ] + a[ j ] 表示前 j-1个数分成 i -1 组，第 j 个数为单独的一组
代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N=1000005;
const int inf=0x3f3f3f;
int a[N],dp[N],ma[N];
int main()
{
int m,n;
while(~scanf("%d%d",&m,&n))
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
memset(dp,0,sizeof(dp));
memset(ma,0,sizeof(ma));
int maxx;
for(int i=1;i<=m;i++)//分成i组
{
maxx=-inf; //记录分成i组的最大和
for(int j=i;j<=n;j++) //前j个数分成i组，至少需要i个数
{
//dp记录前j个数分成i组的最大值,ma[j-1]为前j-1个数分成i-1组的最大和
dp[j]=max(dp[j-1]+a[j],ma[j-1]+a[j]);
ma[j-1]=maxx;//更新前j-1个数分成i组的最大和
maxx=max(maxx,dp[j]);
}
}
printf("%d\n",maxx);//最后的maxx记录的就是n个数分成m组的最大和
}
return 0;
}