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最大子矩阵 hdu1081
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12471 Accepted Submission(s): 5985
Problem Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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.
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 x 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
40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2
Sample Output
15
#include <iostream>#include <algorithm>#include <vector>#include<string.h>using namespace std;int a[55][55];int main(){ int n,m; cin>>n>>m; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { cin>>a[i][j]; a[i][j]+=a[i-1][j]; } } int dp[55]; memset(dp,0,sizeof(dp)); int ans=-0x3fffffff; for(int i=0;i<n-1;i++) { for(int j=i+1;j<n;j++) { for(int k=1;k<=m;k++) { dp[k]=a[j][k-1]-a[i][k-1]; dp[k]=dp[k]+dp[k-1]; if(dp[k]>ans) ans=dp[k]; if(dp[k]<0) dp[k]=0; } } } printf("%d\n",ans); return 0;}
令a[i][k]保存第k列，前i行的和。这样将矩阵通过a[j][k]-a[i][k]压缩成一维，然后求最大字串和。。
最大子矩阵 hdu1081
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