2 seconds
512 megabytes
standard input
standard output
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.
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 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.
5 5
.*.*.
*****
.*.*.
*****
.*.*.
.*.*.
*****
.*.*.
*****
.*.*.
6 7
***.*.*
..*.*.*
*.*.*.*
*.*.*.*
..*...*
*******
***...*
..*...*
..*...*
..*...*
..*...*
*******
4 5
.....
.....
..***
..*..
.....
.....
.....
..... 先转一发官方题解：
http://codeforces.ru/blog/entry/17119?locale=en
525D — Arthur and Walls
To solve this problem we need to observe next fact. If in some square whith size 2 × 2 in given matrix there is exactly one asterisk, we must change it on dot. That is if in matrix from dots and asterisks is not square 2 × 2 in which exactly one asterisk and three dots, then all maximum size of the area from dots connected by sides represent rectangles.
Now solve the problem with help of bfs and this fact. Iterate on all asterisks in given matrix and if only this asterisk contains in some 2 × 2 square, change this asterisk on dot and put this position in queue. Than we need to write standart bfs, in which we will change asterisks on dots in all come out 2 × 2 squares with exactly one asterisk.
Asymptotic behavior of this solution — O(n * m), where n and m sizes of given matrix.
题意：
将 由 '.' 构成的联通区域变成矩形（通过将 '*' 改成 '.' ）
题解：
很不错的一个想法，2*2 的矩形中，如果有一个 '*' 与三个 '.' ，那么这个 '*' 就一定要变成 ‘.' ，然后就bfs 找类似这种 2*2的矩形。
| 10495549 | 2015-03-28 05:12:33 | njczy2010 | D - Arthur and Walls | GNU C++ | Accepted | 732 ms | 27100 KB |
#include <cstdio>
#include <cstring>
#include <stack>
#include <vector>
#include <map>
#include <algorithm>
#include <queue> #define ll long long
int const N = ;
int const M = ;
int const inf = ;
ll const mod = ; using namespace std; int n,m;
char s[N][N];
int dirx[]={-,-,,,,,,-};
int diry[]={,,,,,-,-,-};
int used[N][N]; typedef struct
{
int x;
int y;
}PP; void ini()
{
int i;
for(i=;i<=n;i++){
scanf("%s",s[i]+);
}
memset(used,,sizeof(used));
} int isok(int x,int y)
{
if(x>= && x<=n && y>= && y<=m && s[x][y]=='.'){
return ;
}
else{
return ;
}
} int check(int x,int y)
{
if(x< || x>n || y< || y>m || s[x][y]=='.') return ;
int i;
int d[];
for(i=;i<;i++){
d[i]=isok(x+dirx[i],y+diry[i]);
}
d[]=d[];
// printf(" x=%d y=%d\n",x,y);
// for(i=0;i<=8;i++){
// printf(" i=%d d=%d\n",i,d[i]);
// }
for(i=;i<;i+=){
//printf(" i=%d d=%d\n",i,d[i]);
if(d[i] && d[i+] && d[i+])
return ;
}
return ;
} void solve()
{
queue< PP > que;
PP te,nt;
int i,j;
for(i=;i<=n;i++){
for(j=;j<=m;j++){
if(check(i,j)==){
te.x=i;te.y=j;
//printf(" x=%d y=%d\n",te.x,te.y);
que.push(te);
used[i][j]=;
}
}
} while(que.size()>=){
te=que.front();que.pop();
//printf(" x=%d y=%d\n",te.x,te.y);
s[te.x][te.y]='.';
for(i=;i<;i++){
nt.x=te.x+dirx[i];
nt.y=te.y+diry[i];
if(used[nt.x][nt.y]== && check(nt.x,nt.y)==){
used[nt.x][nt.y]=;
que.push(nt);
}
}
}
} void out()
{
int i;
for(i=;i<=n;i++){
printf("%s\n",s[i]+);
}
} int main()
{
//freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
//scanf("%d",&T);
//for(int cnt=1;cnt<=T;cnt++)
//while(T--)
while(scanf("%d%d",&n,&m)!=EOF)
{
ini();
solve();
out();
}
}