poj 1050(矩阵求和问题dp)

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 44765		Accepted: 23700

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1 8 0 -2

is in the lower left corner:

9 2

-4 1

-1 8

and has a sum of 15.

Input

The
input consists of an N * N array of integers. The input begins with a
single positive integer N on a line by itself, indicating the size of
the square two-dimensional array. This is followed by N^2 integers
separated by whitespace (spaces and newlines). These are the N^2
integers of the array, presented in row-major order. That is, all
numbers in the first row, left to right, then all numbers in the second
row, left to right, etc. N may be as large as 100. The numbers in the
array will be in the range [-127,127].

Output

Output the sum of the maximal sub-rectangle.

Sample Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4  1 -1

8  0 -2

Sample Output

15

Source

Greater New York 2001

 

思路：将其化简成子段求和问题：将第i行到第j行的数累加成一维数列，在用子段求和求解。

　　　

 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <set>
 #define INF 0x3f3f3f3f
 #define INFL 0x3f3f3f3f3f3f3f3f
 #define zero_(x,y) memset(x , y , sizeof(x))
 #define zero(x) memset(x , 0 , sizeof(x))
 #define MAX(x) memset(x , 0x3f ,sizeof(x))
 using namespace std ;
 #define N 505
 typedef long long LL ;
 LL dp[N],a[N][N],s[N][N];
 int main(){
     //freopen("in.txt","r",stdin);
     int n;
     scanf("%d",&n);
     zero(dp);zero(a);zero(s);
     for(int i=;i<=n;i++){
         for(int j=;j<=n;j++){
             scanf("%I64d",&a[i][j]);
         }
     }
     for(int i=;i<=n;i++){
         for(int j=;j<=n;j++){
             s[j][i]=s[j-][i]+a[j][i];

         }
     }

     LL sum=,b=;
     for(int i=;i<=n;i++){
         for(int j=i+;j<=n;j++){
                 b=;
             for(int k=;k<=n;k++){

                 if(b>)
                     b+=(s[j][k]-s[i][k]);
                 else
                     b=s[j][k]-s[i][k];
                 if(b>sum)
                     sum=b;

             }
         }
     }
     printf("%I64d\n",sum);
     return ;
 }