【原】 POJ 2593 Max Sequence 动态规划 解题报告

 

http://poj.org/problem?id=2593

 

Maximum Sequence
求数组两段不重叠的连续子数组的最大和
详见2479

 

Description

Give you N integers a1, a2 ... aN (|ai| <=1000, 1 <= i <= N).

You should output S.

Input

The input will consist of several test cases. For each test case, one integer N (2 <= N <= 100000) is given in the first line. Second line contains N integers. The input is terminated by a single line with N = 0.

Output

For each test of the input, print a line containing S.

Sample Input

5

-5 9 -5 11 20

0

Sample Output

40

 

#include <stdio.h>

#include <iostream>

   3:  

namespace std ;

   5:  

   6:  

int N = 100000 ;

   8:  

int a[N] ;

int maxsofar[2][N] ;

  11:  

void run2593()

  13: {

int n,val ;

int i,j ;

int maxendinghere ;

int max ;

int sum ;

  19:  

, &n ) && n!=0 )

  21:     {

, &a[0] ) ;

  23:         maxendinghere = a[0] ;

  24:         maxsofar[0][0] = a[0] ;

for( i=1 ; i<n ; ++i )

  26:         {

, &a[i] ) ;

  28:             maxendinghere = std::max( a[i] , maxendinghere+a[i] ) ;

  29:             maxsofar[0][i] = std::max( maxsofar[0][i-1] , maxendinghere ) ;

  30:         }

  31:  

  32:         maxendinghere = a[n-1] ;

  33:         maxsofar[1][n-1] = a[n-1] ;

  34:         max = maxsofar[0][n-2] + a[n-1] ;

for( i=n-2 ; i>0 ; --i )

  36:         {

  37:             maxendinghere = std::max( a[i] , a[i]+maxendinghere ) ;

  38:             maxsofar[1][i] = std::max( maxsofar[1][i+1] , maxendinghere ) ;

  39:             sum = maxsofar[0][i-1] + maxsofar[1][i] ;

  40:             max = max>sum ? max : sum ;

  41:         }

 , max ) ;

  43:     }

  44: }