hduoj 1024 Max Sum Plus Plus(最大m子段和)

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24192    Accepted Submission(s): 8289

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 
   
   
   
   
#include <iostream>
#include <algorithm>
#include <map>
#include <cstdio>
#include <string>
#include <cstring>
using namespace std;
int s[1000005];
int dp[1000005];
int temp[1000005];
void DP(int m , int n){
    int i , j , k;
    int max;
    memset(dp , 0 ,sizeof(dp));
    memset(temp , 0 , sizeof(temp));
    for(i = 1 ; i <= m ; i++){//用以个for循环遍历
        max = -999999999;//注意这里的max要为无穷小，分别在求i子段最大和时候进行初始化
        for(j = i ; j <= n ; j++){
            dp[j] = (dp[j-1] + s[j]) > (temp[j-1] + s[j])?(dp[j-1] + s[j]) : (temp[j-1] + s[j]);
            temp[j-1] = max;//这里的temp[j-1]用来存储前面j-1的元素i子段最大值，注意不是temp[j]因为还没判 断是否max<dp[j]；
            if(max < dp[j])
               max = dp[j];//更新max
            }
    }
    cout<<max<<endl;
}
int main(){
    int i , j;
    int n , m;
    while(scanf("%d%d" , &m , &n) != EOF){
        for(i = 1 ; i <= n ; i++)
            scanf("%d" , &s[i]);
        DP(m , n);
    }
    return 0;
}