[CF从零单排#6]263A - Beautiful Matrix

题目来源：http://codeforces.com/problemset/problem/263/A

You've got a 5 × 5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:

Swap two neighboring matrix rows, that is, rows with indexes i and i + 1 for some integer i (1 ≤ i < 5).
Swap two neighboring matrix columns, that is, columns with indexes j and j + 1 for some integer j (1 ≤ j < 5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.

Input
The input consists of five lines, each line contains five integers: the j-th integer in the i-th line of the input represents the element of the matrix that is located on the intersection of the i-th row and the j-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.

Output
Print a single integer — the minimum number of moves needed to make the matrix beautiful.

Examples
input
0 0 0 0 0
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
output
3
input
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
output
1

题目大意：

有一个5*5的矩阵，里面有有24个0，1个1。可以将相邻的两行或两列进行交换。如果1在中心位置（第三行、第三列），我们就称矩阵是美丽的。请问给出一个矩阵，求最少进行多少次交换可以得到美丽的矩阵。

题目分析：

模拟题，首先读取时记录1的位置行x，列y。最少交换次数就是x距离3的长度+y距离3的长度，只需要abs(x-3)+abs(y-3)即可。

参考代码：

#include <bits/stdc++.h>
using namespace std;
int main(){
	int a[6][6], x, y;
	for(int i=1; i<=5; i++)
		for(int j=1; j<=5; j++){
			cin >> a[i][j];
			if(a[i][j]==1)
				x = i, y = j;
		}
	cout << abs(x-3)+abs(y-3);
	return 0;
}