題目大意：給出 n*n 的矩陣，找每隔數字之和最大的子矩陣，輸出最大和。?
解題思路：枚舉矩陣左上和右下的坐標，分別合并子矩陣的每列，使得二維轉化為一維，然后利用連續子序列最大和去做就行。

Time limit ? ?3000 ms

題目描述：

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the subrectangle with the largest sum is referred to as the maximal sub-rectangle.

A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array. 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-hand corner:

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??9 ? ? ?2

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ???4 ? ? ?1

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?1 ? ? ?8

and has the 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 N2 integers separated by white-space (newlines and spaces). These N2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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

The output is 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

題目理解：即在矩形中得到一個小矩形中總和最大數，即遍歷長寬高

#include<iostream> #include<string.h> #define inf 0x3f3f3f3f using namespace std; int n,dp[1010][1010],ans; int main() {while(cin>>n){for(int i=0; i<n; i++)for(int j=0; j<n; j++)cin>>dp[i][j];ans=dp[0][0];for(int i=0; i<n; i++)for(int j=0; j<n; j++)if(j>=i){int sum=0;for(int k=0; k<n; k++){int step=0;for(int u=i; u<=j; u++){step+=dp[u][k];}sum+=step;if(sum>ans)ans=sum;if(sum<0)sum=0;}}cout<<ans<<endl;}return 0; }

?

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