To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14265 Accepted Submission(s): 6775
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
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
Sample Output
15
Source
参考之前的题目吧。。。。。
代码:
#include<bits/stdc++.h> #define ll long long using namespace std; const int maxn = 1e2+7; int n, m[maxn][maxn], sum[maxn][maxn]; void clomn() { memset(sum, 0, sizeof(sum)); for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) sum[i][j] = sum[i - 1][j] + m[i][j]; } int solve() { clomn(); int ans = -10000; for(int r1 = 1; r1 <= n; r1++) for(int r2 = r1; r2 <= n; r2++) { int t = 0; for(int i = 1; i <= n; i++) { t += sum[r2][i] - sum[r1 - 1][i]; ans = max(t, ans); if(t < 0) t = 0; } } return ans; } int main() { //freopen("in.txt", "r", stdin); while(~scanf("%d",&n)) { for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) scanf("%d",&m[i][j]); printf("%d\n", solve()); } return 0; }