Description
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.
Input
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.
Output
Sample Input
5 4 #### #..# #..# #..# ####
Sample Output
2
Hint
#include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; const int MAX = 55; char map1[MAX][MAX]; int flag[MAX][MAX]; int vis[MAX][MAX]; int n,m; int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}}; queue<int>q; int bfs(){ int temp1 = 0; for(int i = 1; i <= n ; i++) for(int j = 1; j <= m ;j++) if(map1[i][j] == '#') temp1++; if(temp1 <= 2) return -1; for(int i = 1; i <= n;i++){ for(int j = 1; j <= m ;j++){ if(map1[i][j] == '#'){ memset(flag,0,sizeof(flag)); int sx = i; int sy = j; map1[sx][sy] = '.'; int dx,dy; for(int w = 0 ; w < 4; w++){ dx = sx + dir[w][0]; dy = sy + dir[w][1]; if ( dx >= 1&&dx <= n && dy >= 1 && dy <= m) break; } while(!q.empty()) q.pop(); q.push(dx); q.push(dy); memset(vis,0,sizeof(vis)); while(!q.empty()){ int x = q.front(); q.pop(); int y = q.front(); q.pop(); flag[x][y] = 1; vis[x][y] = 1; for(int w = 0; w < 4; w++){ int fx = x + dir[w][0]; int fy = y + dir[w][1]; if(fx >= 1 && fx <= n && fy >= 1 && fy <= m &&map1[fx][fy] == '#' && vis[fx][fy] == 0){ q.push(fx); q.push(fy); } } } int k,l; for( k = 1; k <= n; k++){ for( l = 1; l <= m ; l++){ if(map1[k][l] == '#' && flag[k][l] == 0) { return 1; } } } map1[i][j] = '#'; } } } return 2; } int main() { while(~scanf("%d%d",&n,&m)){ getchar(); memset(flag,0,sizeof(flag)); memset(map1,0,sizeof(map1)); for(int i = 1; i <= n; i++){ for(int j = 1; j <= m ; j++) scanf("%c",&map1[i][j]); getchar(); } /* for(int i = 1; i <= n; i ++){ for(int j = 1; j <= m ; j++){ printf("%c",map1[i][j]); } printf(" "); } */ printf("%d ",bfs()); } return 0; }
并查集写法：
#include<cstdio> #include<cstring> const int MAX = 2800; int p[MAX]; int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}}; int n,m; char map1[55][55]; int find(int x){ return x == p[x] ? x : p[x] = find(p[x]); } int solve() { int temp = 0; for(int i = 1 ;i <= n; i++) for(int j = 1; j <= m; j++) if(map1[i][j] == '#') temp ++; if(temp <= 2) return -1; for(int i = 1; i <= n ;i++){ for(int j = 1; j <= m ;j++){ if(map1[i][j] == '#'){ for(int k = 1; k <= n*m ;k++) p[k] = k; map1[i][j] = '.'; for(int k = 1; k <= n; k++){ for(int l = 1; l <= m; l++){ if(map1[k][l] == '#'){ for(int d = 0; d < 4 ; d++){ int x = k + dir[d][0]; int y = l + dir[d][1]; if( x >= 1 && x <= n && y >= 1 && y <= n && map1[x][y] == '#') p[find(k*m+l)] = find(x*m+y); } } } } int num = 0; for(int k = 1; k <= n*m;k++){ if(find(k) == k) num++; } if(num > 1) return 1; map1[i][j] = '#'; } } } return 2; } int main() { while(~scanf("%d%d",&n,&m)){ memset(map1,0,sizeof(map1)); memset(p,0,sizeof(p)); getchar(); for(int i = 1; i <= n ; i++){ for(int j = 1; j <= m; j++){ scanf("%c",&map1[i][j]); } getchar(); } /* for(int i = 1; i <= n ; i++) for(int j = 1; j <= m;j++) printf("%c",map1[i][j]); */ printf("%d ",solve()); } return 0; }