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
其实就是暴力一个最大字段和。。 本身最大字段和是求一行的 这个要求一个最大的矩形
那么你就枚举每一个可能出现的行…、 比如1到3，1到4全都叠加完了成为一个新的行 对新的行最大字段和就可以了..
之前我竟然往区间dp上想.. 有点傻逼
#include<iostream> #include<algorithm> #include<cstdio> using namespace std; int tu[101][101]; int main() { int n; cin >> n; int da = -100000000; for (int a = 1; a <= n; a++) { int tem = -10000000; for (int b = 1; b <= n; b++) { scanf("%d", &tu[a][b]); if (tem > 0)tem += tu[a][b]; else tem = tu[a][b]; da = max(da, tem); } } for (int a = 1; a <= n; a++) { for (int b = a + 1; b <= n; b++) { int tem = -1000000; for (int c = 1; c <= n; c++) { tu[a][c] += tu[b][c]; if (tem > 0)tem += tu[a][c]; else tem = tu[a][c]; da = max(tem, da); } } } cout << da; }