Pku acm 1050 to the Max [Dynamic Planning]

The meaning of the question is very simple. Find its sub-matrix in a matrix to maximize the sum of the sub-Matrix Values. In fact, it is the promotion of the largest sub-segment and problem in two-dimensional space. Let's talk about the one-dimensional situation first: It has arrays A0, A1... An. Find the consecutive sub-segments to maximize the sum of them. Assume that for sub-segments: 9 2-16 2 temp [I], it indicates the maximum sub-segment and in the sub-segment ending with AI. If temp [I] is known
The [I + 1] method is:

If temp [I]> 0 temp [I + 1] = temp [I] + AI (add AI to the previous child segment ), otherwise, temp [I + 1] = AI (without the preceding sub-segment), that is, the state transition equation:

Temp [I] = (temp [I-1]> 0? Temp [I-1]: 0) + Buf [I];

For the example temp: 9 11-5 2, then the largest sub-segment in temp [] is the largest of the one-dimensional sequence. Evaluate the functions of the largest subsegment and in one dimension:

Int getmax (INT Buf [100], int N)

{

Int temp [101], max = N * (-127 );

Memset (temp, 0, 4 * (n + 1 ));

For (INT I = 1; I <= N; I ++)

{

Temp [I] = (temp [I-1]> 0? Temp [I-1]: 0) + Buf [I];

If (max <temp [I])

Max = temp [I];

}

Return Max;

}

The following is an example of two-dimensional scaling:

0-2-7 0

9 2-6 2

-4 1-4 7

-1 8 0-2

We use I j to indicate the start and end rows respectively, and traverse all possibilities:

For (I = 1; I <= N; I ++)

For (j = I; j <= N; j ++ ){}

In one of the cases, I = 2 J = 4, which is equivalent to selecting 2 3 4 and 3 rows, so that the combination of those columns can obtain the maximum value, because it is always 2 3 4 rows, therefore, we can bind these three lines to 4 (9-4-1), 11 (8 + 2 + 1 ), -10 (-6-4 + 0), 7 (7 + 2-2) Maximum child segment and, OK, the problem is successfully converted to one-dimensional situation!

// Acm pku 1050 to the max by suqiang @ neuq & jlu // For more detailed Problem Solving reports please visit my blog: http://blog.csdn.net/china8848#include <stdio. h> # include <memory. h> // calculate the maximum sub-segment of the one-dimensional array and INT getmax (INT Buf [100], int N) {int temp [101], max = N * (-127); memset (temp, 0, 4 * (n + 1); For (INT I = 1; I <= N; I ++) {temp [I] = (temp [I-1]> 0? Temp [I-1]: 0) + Buf [I]; If (max <temp [I]) max = temp [I];} return Max;} int main (void) {int N, num [101] [101], I, J, K, Max, temp [101]; scanf ("% d", & N ); for (I = 1; I <= N; I ++) for (j = 1; j <= N; j ++) scanf ("% d ", & num [I] [J]); max =-127 * n; for (I = 1; I <= N; I ++) {memset (temp, 0, 4 * (n + 1); For (j = I; j <= N; j ++) {for (k = 1; k <= N; k ++) {temp [k] + = num [J] [k];} // convert two-dimensional data to one-dimensional int this_max = getmax (temp, n); If (this_max> MAX) max = this_max ;}} printf ("% d/N", max); return 1 ;}