SGU 168.Matrix

时间限制：0.5s

空间限制：15M

题意：

      给出一个N*M的矩阵A，计算矩阵B，满足B[i][j]=min{ A[x][y]:(y>=j) and ( x>=i+j-y )}

 

 

 

 

 

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Solution : 

             

              如图方式从右下角遍历矩阵，那么可令B[i][j]=min(A[i][j],B[i-1][j],B[i][j+1],B[i-1][j+1])

              动态规划即可。

              时间复杂度O（n*m)

 

code

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

int A[1009][1009], B[1009][1009];
int n, m;

int main() {
	scanf ("%d %d", &n, &m);
	memset (B, 0x3f, sizeof B);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++) scanf ("%d", &A[i][j]);
	int x = n, y = m, i = n, j = m;
	while (x >= 1 || y >= 1) {
		while (i <= n && j >= 1) {
			B[i][j]  = min (min (A[i][j], B[i][j + 1]), min (B[i + 1][j], B[i - 1][j + 1]) );
			i++, j--;
		}
		if (i > n || j < 1) {
			i = --x, j = y;
			if (i <= 0) i = 1, j = --y;
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++)
			printf ("%d ", B[i][j]);
		putchar (10);
	}
	return 0;
}