hdu---1024Max 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): 15898    Accepted Submission(s): 5171

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_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 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

Hint

Huge input, scanf and dynamic programming is recommended.

 

Author

JGShining（极光炫影）

 

 题意：

   给定一段连续数字串，要你求出m个连续子串的最大值。

 比如

   下面这个样列

    2  6 -1 4 -2 3 -2 3

    给出一个长度为6的连续数字串，要你求出这个串中连续的m=2个子串的子串之和.......

   

对样列的一个数值分析

aa	-1	4	-2	3	-2	3
第一遍 maxc	-1	4	4	4	5
dp	-1	4	2	5	3	6
第二遍 maxc	-1	3	3	7	7
dp	-1	3	2	7	5	8

代码

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int a[1000001],dp[1000001],max1[1000001];
 int max(int x,int y){
    return x>y?x:y;
}
  int main(){
    int i,j,n,m,temp;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
         dp[0]=0;
        for(i=1;i<=n;i++)
        {
          scanf("%d",&a[i]);
          dp[i]=0;
          max1[i]=0;
         }
        max1[0]=0;
        for(i=1;i<=m;i++){
            temp=-0x3f3f3f3f;
            for(j=i;j<=n;j++){
                dp[j]=max(dp[j-1]+a[j],max1[j-1]+a[j]);
                max1[j-1]=temp;
                temp=max(temp,dp[j]);
            }
        }
        printf("%d
",temp);
    }
    return 0;
  }

 优化后的代码：

/*hdu 1024 @coder Gxjun*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
const int maxn=1000005;
int aa[maxn],dp[maxn],maxc[maxn];
int max(int a,int b){
  return a>b?a:b;
}
int main()
{
    int n,m,i,j,temp;
    while(scanf("%d%d",&m,&n)!=EOF){
        memset(maxc,0,sizeof(int)*(n+1));
        memset(dp,0,sizeof(int)*(n+1));
       for(i=1;i<=n;i++)
          scanf("%d",&aa[i]);
       for(i=1;i<=m;i++){
          temp=-0x3f3f3f3f;
       for(j=i;j<=n;j++){
          dp[j]=max(dp[j-1],maxc[j-1])+aa[j];
          maxc[j-1]=temp;
          temp=max(temp,dp[j]);
          }
        }
        printf("%d
",temp);
    }
}