Codeforces Round #375 (Div. 2) D. Lakes in Berland 贪心

D. Lakes in Berland

题目连接：

http://codeforces.com/contest/723/problem/D

Description

The map of Berland is a rectangle of the size n × m, which consists of cells of size 1 × 1. Each cell is either land or water. The map is surrounded by the ocean.

Lakes are the maximal regions of water cells, connected by sides, which are not connected with the ocean. Formally, lake is a set of water cells, such that it's possible to get from any cell of the set to any other without leaving the set and moving only to cells adjacent by the side, none of them is located on the border of the rectangle, and it's impossible to add one more water cell to the set such that it will be connected with any other cell.

You task is to fill up with the earth the minimum number of water cells so that there will be exactly k lakes in Berland. Note that the initial number of lakes on the map is not less than k.

Input

The first line of the input contains three integers n, m and k (1 ≤ n, m ≤ 50, 0 ≤ k ≤ 50) — the sizes of the map and the number of lakes which should be left on the map.

The next n lines contain m characters each — the description of the map. Each of the characters is either '.' (it means that the corresponding cell is water) or '*' (it means that the corresponding cell is land).

It is guaranteed that the map contain at least k lakes.

Output

In the first line print the minimum number of cells which should be transformed from water to land.

In the next n lines print m symbols — the map after the changes. The format must strictly follow the format of the map in the input data (there is no need to print the size of the map). If there are several answers, print any of them.

It is guaranteed that the answer exists on the given data.

Sample Input

5 4 1

---

..

---

.

..*

Sample Output

1

---

..

---

---

..**

Hint

题意

给你一个n/*m的矩阵，然后你们有不少于k条河流，然后你需要使得一些河流变成陆地，使得河流的数量恰好等于k，问你至少填多少个水。

河流的定义是水塘，且不与外界相连的地方。

题解：

直接dfs搜出每一条河流，然后贪心排序，从小到大去填满就好了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 55;
int n,m,k;
string s[maxn];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int vis[maxn][maxn];
int sp[maxn*maxn],cnt,area,flag;
void dfs(int x,int y)
{
    area++;
	vis[x][y]=1;
	if(x==0||x==n-1||y==0||y==m-1)flag=1;
	for(int i=0;i<4;i++)
	{
		int nx=x+dx[i];
		int ny=y+dy[i];
		if(nx<0||nx>=n)continue;
		if(ny<0||ny>=m)continue;
		if(s[nx][ny]=='*')continue;
		if(vis[nx][ny])continue;
		dfs(nx,ny);
	}
}
struct node
{
	int a,b,c;
}t[5000];
bool cmp(node a,node b)
{
	return a.a<b.a;
}
void dfs2(int x,int y)
{
	s[x][y]='*';
	for(int i=0;i<4;i++)
	{
		int nx=x+dx[i];
		int ny=y+dy[i];
		if(nx<0||nx>=n)continue;
		if(ny<0||ny>=m)continue;
		if(s[nx][ny]=='*')continue;
		dfs2(nx,ny);
	}
}
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for(int i=0;i<n;i++)cin>>s[i];
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			if(!vis[i][j]&&s[i][j]=='.')
			{
				area=0;
				flag=0;
				dfs(i,j);
				if(flag==1)continue;
				t[cnt].a=area,t[cnt].b=i,t[cnt].c=j;
				cnt++;
			}
		}
	}
	sort(t,t+cnt,cmp);
	int aaans = 0;
	for(int i=0;i<cnt-k;i++)
	{
		aaans+=t[i].a;
		dfs2(t[i].b,t[i].c);
	}
	cout<<aaans<<endl;
	for(int i=0;i<n;i++)
		cout<<s[i]<<endl;
}