POJ 1050 To the Max(DP_最大字段和)

题目链接DescriptionGiven 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 thesub-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 09 2 -6 2-4 1 -4 1-1 8 0 -2is in the lower left corner:9 2-4 1-1 8and has a sum of 15.InputThe 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].OutputOutput 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

求最大子矩阵和解题思路：a11 a12 a13a21 a22 a23a31 a32 a33可以先求出a11, a12, a13的最大子段和，然后再求出a11+a21, a12+a22, a13+a23的最大子段和，接着求a11+a21+a31, a12+a22+a32, a13+a23+a33的最大子段和，求完以第一行开头的，再接着求以第二行开头的，以此类推求出所有的最大子段和，求出他们中最大的就是解。代码如下：

/*_ooOoo_o8888888o88" . "88(| -_- |)O\ = /O____/`---'\____.   ' \\| |// `./ \\||| : |||// \/ _||||| -:- |||||- \| | \\\ - /// | || \_| ''\---/'' | |\ .-\__ `-` ___/-. /___`. .' /--.--\ `. . __."" '< `.___\_<|>_/___.' >'"".| | : `- \`.;`\ _ /`;.`/ - ` : | |\ \ `-. \_ __\ /__ _/ .-` / /======`-.____`-.___\_____/___.-`____.-'======`=---='.............................................*/#include <map>#include <queue>#include <cmath>#include <cstdio>#include <string>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define pi 3.1415926#define INF 123456789using namespace std;int main(){int n, i, j, a[105][105], b[105], q, t, Max;while(~scanf("%d", &n)){for(i = 0; i < n; i ++){for(j = 0; j < n; j ++)scanf("%d", &a[i][j]);}for(i = 0, Max = -INF; i < n; i ++){memset(b, 0, sizeof(b));for(j = i; j < n; j ++){for(q = 0; q < n; q ++)b[q] += a[j][q];for(q = 0, t = 0; q < n; q ++){t += b[q];if(t > Max) Max = t;if(t < 0) t = 0;}}}printf("%d\n", Max);}return 0;}