poj1050动态规划求最大矩阵和.docx

To the MaxTime Limit:1000MSMemory Limit:10000KTotal Submissions:32984Accepted:17276DescriptionGiven 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 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 N2 integers separated by whitespace (spaces and newlines). These are the N2 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 Input40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2Sample Output15当然是dp了只是如何去分解每一步，成了问题。其实把它转化成求一位数组的最大子序列问题就行代码如下：#include#include#includeusing namespace std;const int MAX=105;int aMAXMAX;int tMAX;int n;int get() /最大和子序列函数 int m=0,te=0; for(int i=1;im) m=te; if(ten) for(int i=1;i=n;+i) for(int j=1;j=n;+j) scanf(%d,&aij); int num=0; for(int i=1;i=n;+i) / i行开始 memset(t,0,sizeof(t); for(int j=i;j=n;+j) /j行结束