杭电1024Max Sum Plus Plus

A - Max Sum Plus Plus

Time 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 32 6 -1 4 -2 3 -2 3 

 

Sample Output

68 

Hint

 Huge input, scanf and dynamic programming is recommended.     求n个元素m段连续子段和，dp[i][j]：前j个数i段的最大和。dp[i][j]=max(dp[i][j-1],dp[i-1][k])+a[i][j];

若第j个数单独构成一段，则求出前j-1个数构成i-1段的最大和，或者求出前j-1个数构成i段的最大和     

 #include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f
using namespace std;
int dp[1000010];
int pre[1000010];
int a[1000010];
int main()
{
int m,n,i,j,ma;
while(~scanf("%d%d",&m,&n))
{
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
memset(dp,0,sizeof(dp));
memset(pre,0,sizeof(pre));
for(i=1;i<=m;i++)
{
ma=-INF;
for(j=i;j<=n;j++)
{
dp[j]=max(dp[j-1],pre[j-1])+a[j];
pre[j-1]=ma;
ma=max(ma,dp[j]);
}
}
printf("%d\n",ma);
}
return 0;
}