POJ 1050-To The Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 41968		Accepted: 22303

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

小结：

我开始是确定两个点作为一个区域的对角线的两个点，然后通过逐次通过面积的加减来求出相应的解，然而写着写着就发现不对了，感觉好像时间复杂度为n*4，当n为100的时候感觉就有TLE的危险，果断就放弃了（说实话，后面感觉好像这样也可以的...），然后百思不得其解，就参考了前辈的做法....顿时感觉自己好没用....

以下是AC代码：

#include<stdio.h>#include<string.h>#include<stdlib.h>int num[110][110];int temp[110];int n;int find(int *a){int sum=0;int max=0;for(int i=1;i<=n;i++){sum+=a[i];if(sum<0)sum=0;max=max>sum?max:sum;}return max;}int main(){int max;while(~scanf("%d",&n)){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){scanf("%d",&num[i][j]);}}for(int i=1;i<=n;i++){memset(temp,0,sizeof(temp));for(int j=i;j<=n;j++){for(int k=1;k<=n;k++)temp[k]+=num[j][k];int t=find(temp);max=max>t?max:t;}}printf("%d\n",max);}return 0;}