hdu---1024Max Sum Plus Plus(动态规划)

时间：2014-06-02 14:59:54 收藏：0 阅读：247

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）

代码：

 1 #include<iostream>
 2 #include<string.h>
 3 #include<stdio.h>
 4 using namespace std;
 5 int a[1000001],dp[1000001],max1[1000001];
 6  int max(int x,int y){
 7     return x>y?x:y;
 8 }
 9   int main(){
10     int i,j,n,m,temp;
11     while(scanf("%d%d",&m,&n)!=EOF)
12     {
13          dp[0]=0;
14         for(i=1;i<=n;i++)
15         {
16           scanf("%d",&a[i]);
17           dp[i]=0;
18           max1[i]=0;
19          }
20         max1[0]=0;
21         for(i=1;i<=m;i++){
22             temp=-0x3f3f3f3f;
23             for(j=i;j<=n;j++){
24                 dp[j]=max(dp[j-1]+a[j],max1[j-1]+a[j]);
25                 max1[j-1]=temp;
26                 temp=max(temp,dp[j]);
27             }
28         }
29         printf("%d\n",temp);
30     }
31     return 0;
32   }

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