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hdu---1024Max Sum Plus Plus(动态规划)

时间:2014-06-02 14:59:54      阅读:247      评论:0      收藏:0      [点我收藏+]

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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 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^
 

 

Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
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:

bubuko.com,布布扣
 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   }
View Code

 

hdu---1024Max Sum Plus Plus(动态规划),布布扣,bubuko.com

hdu---1024Max Sum Plus Plus(动态规划)

标签:des   c   style   class   blog   code   

原文地址:http://www.cnblogs.com/gongxijun/p/3764266.html

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