You've gotten an n?×?m sheet of squared paper. Some of its squares are painted. Let's mark the set of all painted squares as A. Set A is connected. Your task is to find the minimum number of squares that we can delete from set A to make it not connected.
A set of painted squares is called connected, if for every two squares a and b from this set there is a sequence of squares from the set, beginning in a and ending in b, such that in this sequence any square, except for the last one, shares a common side with the square that follows next in the sequence. An empty set and a set consisting of exactly one square are connected by definition.
The first input line contains two space-separated integers n and m (1?≤?n,?m?≤?50) — the sizes of the sheet of paper.
Each of the next n lines contains m characters — the description of the sheet of paper: the j-th character of the i-th line equals either "#", if the corresponding square is painted (belongs to set A), or equals "." if the corresponding square is not painted (does not belong to set A). It is guaranteed that the set of all painted squares A is connected and isn't empty.
On the first line print the minimum number of squares that need to be deleted to make set A not connected. If it is impossible, print -1.
5 4 #### #..# #..# #..# ####
2
5 5 ##### #...# ##### #...# #####
2
In the first sample you can delete any two squares that do not share a side. After that the set of painted squares is not connected anymore.
The note to the second sample is shown on the figure below. To the left there is a picture of the initial set of squares. To the right there is a set with deleted squares. The deleted squares are marked with crosses.
题解：
给你个图为一个#号的连通块，问你至少去掉几个#可以使图不连通
看起来很难但是找找规律可以发现无论是什么图的答案都是-1,1,或者2，因为去掉最多的情况把图形的角的方格两边连通块去掉就是用2个
那么-1的情况就是#小于3个的时候
枚举所有#号位置，如果去掉该位置连通块是2个以上就直接输出1，否则输出2
代码：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#include<stdio.h>
using namespace std;
#define ll long long
int dirx[4]={0,0,-1,1};
int diry[4]={1,-1,0,0};
char s[55][55];
int vis[55][55];
int n,m,ans;
void dfs(int x,int y)
{ans++;int xx,yy,i;for(i=0;i<4;i++){xx=dirx[i]+x;yy=diry[i]+y;if(xx>=0&&yy>=0&&xx<n&&yy<m&&!vis[xx][yy]&&s[xx][yy]=='#'){vis[xx][yy]=1;dfs(xx,yy);}}
}
int main()
{int i,j,num=0;scanf("%d%d",&n,&m);for(i=0;i<n;i++){scanf("%s",s[i]);for(j=0;j<m;j++){if(s[i][j]=='#'){num++;}}}if(num<3){printf("-1\n");return 0;}for(i=0;i<n;i++){for(j=0;j<m;j++){if(s[i][j]=='#'){memset(vis,0,sizeof(vis));vis[i][j]=1;ans=0;for(int p=0;p<n;p++){for(int q=0;q<m;q++){if(s[p][q]=='#'&&!vis[p][q]){vis[p][q]=1;dfs(p,q);goto loop;}}}loop:;if(ans!=num-1){printf("1\n");return 0;}}}}printf("2\n");return 0;
}