Codeforces Round #209 (Div. 2) A.Table

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Simon has a rectangular table consisting of n rows and m columns. Simon numbered the rows of the table from top to bottom starting from one and the columns — from left to right starting from one. We’ll represent the cell on the x-th row and the y-th column as a pair of numbers (x, y). The table corners are cells: (1, 1), (n, 1), (1, m), (n, m).

Simon thinks that some cells in this table are good. Besides, it’s known that no good cell is the corner of the table.

Initially, all cells of the table are colorless. Simon wants to color all cells of his table. In one move, he can choose any good cell of table (x1, y1), an arbitrary corner of the table (x2, y2) and color all cells of the table (p, q), which meet both inequations: min(x1, x2) ≤ p ≤ max(x1, x2), min(y1, y2) ≤ q ≤ max(y1, y2).

Help Simon! Find the minimum number of operations needed to color all cells of the table. Note that you can color one cell multiple times.

Input
The first line contains exactly two integers n, m (3 ≤ n, m ≤ 50).

Next n lines contain the description of the table cells. Specifically, the i-th line contains m space-separated integers ai1, ai2, …, aim. If aij equals zero, then cell (i, j) isn’t good. Otherwise aij equals one. It is guaranteed that at least one cell is good. It is guaranteed that no good cell is a corner.

Output
Print a single number — the minimum number of operations Simon needs to carry out his idea.

Examples
input
3 3
0 0 0
0 1 0
0 0 0
output
4
input
4 3
0 0 0
0 0 1
1 0 0
0 0 0
output
2
Note
In the first sample, the sequence of operations can be like this:

For the first time you need to choose cell (2, 2) and corner (1, 1).
For the second time you need to choose cell (2, 2) and corner (3, 3).
For the third time you need to choose cell (2, 2) and corner (3, 1).
For the fourth time you need to choose cell (2, 2) and corner (1, 3).
In the second sample the sequence of operations can be like this:

For the first time you need to choose cell (3, 1) and corner (4, 3).
For the second time you need to choose cell (2, 3) and corner (1, 1).

题目大意：
给你一个n×m的方格，其中某些方格为1，先要将方格进行染色，染色方式为将任意一个角到某个1进行连线，以这条线作为对角线的方格被染色，问最少几次可以将这个方格染满

思路：
不必考虑所有为1的方格，若四条边上有方格为1，则需要2次，否则需要4次

代码：

#include<cstdio>#include<iostream>#include<cstring>using namespace std;int map[55][55];int main(){    int n,m;    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            scanf("%d",&map[i][j]);        }    }    int flag1=0;    int flag2=0;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            if(map[i][j]==1)            {                if(i==1||i==n||j==1||j==m)                    flag2=1;                else                    flag1=1;            }        }    }    if(flag2==1)        printf("2\n");    else        printf("4\n");    return 0;}