POJ_1050_To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49811		Accepted: 26400

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

 

• 最大子矩阵和
• 是一维的最大子串和的二维扩展
• 那我们把每列做一个前缀和，O1得到从i行到j行单列的和，之后用最大子串和dp求解就行

 #include <iostream>
 #include <string>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <climits>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long           LL ;
 typedef unsigned long long ULL ;
 const int    maxn = 1e2 +    ;
 const int    inf  = 0x3f3f3f3f ;
 const int    npos = -         ;
 const int    mod  = 1e9 +     ;
 const int    mxx  =  +     ;
 const double eps  = 1e-       ;
 const double PI   = acos(-1.0) ;

 int n, ans, a[maxn][maxn], b[maxn], c[maxn][maxn];
 int main(){
     // freopen("in.txt","r",stdin);
     // freopen("out.txt","w",stdout);
     while(~scanf("%d",&n)){
         ans=-inf;
         memset(c,,sizeof(c));
         for(int i=;i<=n;i++)
             for(int j=;j<=n;j++){
                 scanf("%d",&a[i][j]);
                 c[i][j]=c[i-][j]+a[i][j];
             }
         for(int i=;i<=n;i++){
             for(int j=i;j<=n;j++){
                 b[]=;
                 for(int k=;k<=n;k++){
                     if(b[k-]>=){
                         b[k]=b[k-]+c[j][k]-c[i-][k];
                     }else{
                         b[k]=c[j][k]-c[i-][k];
                     }
                     ans=max(ans,b[k]);
                 }
             }
         }
         printf("%d\n",ans);
     }
     return ;
 }