HDU OJ 1081 To The Max
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一开始做的时候没有思路,不知道该怎么做,感觉暴力肯定会超时的。。没敢做,后来看了题解感觉方法不错,就学习了。。。
参考博客 http://blog.csdn.net/acm_davidcn/article/details/5834454 感谢这位大神。
这题的意思就是找到子矩阵中最大的和值。
这是学习的解题思路:DP 求最大和子矩阵 , 可以用最大和连续子序列的思路解 .
首先 , 读数据的时候 a[i][j] 为前 i 行数据的第 j 个数的和 , 当求第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵的时候 , 就以 p 行和 q 行之间的同一列的数字的和作为一个数列的元素 , 然后 DP 求这个数列的最大和连续子序列 , 就是第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵 .
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6060 Accepted Submission(s): 28
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.
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
参考博客 http://blog.csdn.net/acm_davidcn/article/details/5834454 感谢这位大神。
这题的意思就是找到子矩阵中最大的和值。
这是学习的解题思路:DP 求最大和子矩阵 , 可以用最大和连续子序列的思路解 .
首先 , 读数据的时候 a[i][j] 为前 i 行数据的第 j 个数的和 , 当求第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵的时候 , 就以 p 行和 q 行之间的同一列的数字的和作为一个数列的元素 , 然后 DP 求这个数列的最大和连续子序列 , 就是第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵 .
To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6060 Accepted Submission(s): 28
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.
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
#include <stdio.h>#include <string.h>int main(){ int dp[105][105]; int i, j, k, t, max, sum, n; while(~scanf("%d",&n)) { memset(dp , 0 , sizeof(dp)); for(i = 1; i <= n; i++) for(j = 1; j <= n; j++) { scanf("%d",&dp[i][j]); dp[i][j] += dp[i-1][j]; } max = 0; for(i = 1; i <= n; i++) { for(j = i; j <= n; j++) { sum = 0; for(k = 1; k <= n; k++) { t = dp[j][k] - dp[i-1][k]; sum+= t; if(sum > max) max = sum; if(sum < 0) sum = 0; } } } printf("%d\n",max); } return 0;}
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