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\n",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=;
 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,,sizeof(int)*(n+));
         memset(dp,,sizeof(int)*(n+));
        for(i=;i<=n;i++)
           scanf("%d",&aa[i]);
        for(i=;i<=m;i++){
           temp=-0x3f3f3f3f;
        for(j=i;j<=n;j++){
           dp[j]=max(dp[j-],maxc[j-])+aa[j];
           maxc[j-]=temp;
           temp=max(temp,dp[j]);
           }
         }
         printf("%d\n",temp);
     }
 }