题目链接:http://poj.org/problem?id=1050
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.
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
求最大子矩阵元素和,从数据量上看,可以发现非常小,不难猜想是一道比较暴力的题目。
听说是DP?我也不知道求矩阵前缀和算不算DP……
我们用dp[i][j]表示这个矩阵前i行,前j列所有元素的和。
那么,我们求任何一个子矩阵 [x1,y1][x2,y2] ,就用dp[x2][y2]-dp[x2][y1]-dp[x1][y2]+dp[x1 - 1][y1 - 1]表示即可;
然后,直接四重循环枚举x1,x2,y1,y2就可得到答案。
1 #include<cstdio> 2 #include<cstring> 3 #define MAXN 105 4 #define INF 0x3f3f3f3f 5 int n,dp[MAXN][MAXN]; 6 int main() 7 { 8 while(scanf("%d",&n)!=EOF) 9 { 10 memset(dp,0,sizeof(dp)); 11 for(int i=1;i<=n;i++) 12 { 13 for(int j=1;j<=n;j++) 14 { 15 int tmp; 16 scanf("%d",&tmp); 17 dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+tmp; 18 } 19 } 20 int ans=-INF; 21 for(int i=1;i<=n;i++) 22 { 23 for(int j=1;j<=n;j++) 24 { 25 for(int ii=i;ii<=n;ii++) 26 { 27 for(int jj=j;jj<=n;jj++) 28 { 29 int now=dp[ii][jj]-dp[ii][j-1]-dp[i-1][jj]+dp[i-1][j-1]; 30 if(ans<now) ans=now; 31 } 32 } 33 } 34 } 35 printf("%d ",ans); 36 } 37 }