Codeforces #297( div2) D. Arthur and Walls ( DFS

D. Arthur and Walls

Description

Finally it is a day when Arthur has enough money for buying an apartment. He found a great option close to the center of the city with a nice price.

Plan of the apartment found by Arthur looks like a rectangle n × m consisting of squares of size 1 × 1. Each of those squares contains either a wall (such square is denoted by a symbol “*” on the plan) or a free space (such square is denoted on the plan by a symbol “.”).

Room in an apartment is a maximal connected area consisting of free squares. Squares are considered adjacent if they share a common side.

The old Arthur dream is to live in an apartment where all rooms are rectangles. He asks you to calculate minimum number of walls you need to remove in order to achieve this goal. After removing a wall from a square it becomes a free square. While removing the walls it is possible that some rooms unite into a single one.

Input

The first line of the input contains two integers n, m (1 ≤ n, m ≤ 2000) denoting the size of the Arthur apartments.

Following n lines each contain m symbols — the plan of the apartment.

If the cell is denoted by a symbol “*” then it contains a wall.

If the cell is denoted by a symbol “.” then it this cell is free from walls and also this cell is contained in some of the rooms.

Output

Output n rows each consisting of m symbols that show how the Arthur apartment plan should look like after deleting the minimum number of walls in order to make each room (maximum connected area free from walls) be a rectangle.

If there are several possible answers, output any of them.

Sample Input

5 5.*.*.*****.*.*.*****.*.*.

Sample Output

.*.*.*****.*.*.*****.*.*.

题意

给一个n∗m的的图形，里面有“∗”“.”两种，问怎么将”.”全部变成矩形
很有思维的DFS我们知道 矩形最小是 2∗2那么 我们可以遍历每一个这样的矩形 当里面有一个“*”就变为“.”然后 又影响了周围的八个 继续对八个遍历

AC代码

#include <bits/stdc++.h>using namespace std;#define LL long long#define CLR(a,b) memset(a,(b),sizeof(a))const int MAXN = 2e3+10;char mps[MAXN][MAXN];int dx[] = {0,1,1,0};int dy[] = {1,1,0,0};int n, m;void dfs(int x, int y){    if(x<0 || x>=n-1 || y<0 || y>=m-1) return;    int res = 0;    int fx, fy;    for(int i = 0; i < 4; i++) {        int xx = x+dx[i];        int yy = y+dy[i];        if(mps[xx][yy] == '*') {            res++;            fx = xx; fy = yy;        }    }    if(res == 1) {        mps[fx][fy] ='.';        for(int i = -1; i <= 1; i++) {            for(int j = -1; j <= 1; j++) {                 dfs(fx+i,fy+j);            }        }    }}int main(){    scanf("%d%d",&n,&m);    for(int i = 0; i < n; i++)        scanf("%s",mps[i]);    for(int i = 0; i < n-1; i++) {        for(int j = 0;j < m-1; j++) {            dfs(i,j);        }    }    for(int i = 0; i < n; i++)        printf("%s\n",mps[i]);return 0;}