F - Lakes in Berland(BFS)
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 klakes 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.
Example
5 4 1
****
*..*
****
**.*
..**
1
****
*..*
****
****
..**
3 3 0
***
*.*
***
1
***
***
***
Note
In the first example there are only two lakes — the first consists of the cells (2, 2) and (2, 3), the second consists of the cell (4, 3). It is profitable to cover the second lake because it is smaller. Pay attention that the area of water in the lower left corner is not a lake because this area share a border with the ocean.
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstring>
5 #include <map>
6 using namespace std;
7 typedef long long ll;
8 const int N=55;
9 inline int read(){
10 char c=getchar();int x=0,f=1;
11 while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
12 while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
13 return x*f;
14 }
15 int n,m,k;
16 char g[N][N];
17 int dx[4]={1,-1,0,0},dy[4]={0,0,1,-1};
18 int vis[N][N],num[N*N],cc=0;
19 struct lakes{
20 int size,id;
21 }lake[N*N];
22 int cnt=0;
23 bool cmp(lakes &a,lakes &b){
24 return a.size<b.size;
25 }
26 void dfs(int x,int y,int id){//printf("dfs %d %d %d\n",x,y,id);
27 vis[x][y]=id;num[id]++;
28 for(int i=0;i<4;i++){
29 int nx=x+dx[i],ny=y+dy[i];
30 if(nx<1||nx>n||ny<1||ny>m) continue;
31 if(g[nx][ny]=='*'||vis[nx][ny]) continue;
32 dfs(nx,ny,id);
33 }
34 }
35 int ans=0;
36 void fil(int x,int y,int id){//printf("fil %d %d %d\n",x,y,id);
37 g[x][y]='*';ans++;
38 for(int i=0;i<4;i++){
39 int nx=x+dx[i],ny=y+dy[i];
40 if(nx<1||nx>n||ny<1||ny>m) continue;
41 if(vis[nx][ny]==id&&g[nx][ny]=='.') fil(nx,ny,id);
42 }
43 }
44 int main(){
45 n=read();m=read();k=read();
46 for(int i=1;i<=n;i++){
47 scanf("%s",g[i]);
48 for(int j=m;j>=1;j--) g[i][j]=g[i][j-1];
49 }
50
51 for(int i=1;i<=n;i++){
52 if(!vis[i][1]&&g[i][1]=='.') dfs(i,1,++cc);
53 if(!vis[i][m]&&g[i][m]=='.') dfs(i,m,++cc);
54 }
55 for(int j=1;j<=m;j++){
56 if(!vis[1][j]&&g[1][j]=='.') dfs(1,j,++cc);
57 if(!vis[n][j]&&g[n][j]=='.') dfs(n,j,++cc);
58 }
59 int sea=cc;
60 for(int i=1;i<=n;i++)
61 for(int j=1;j<=m;j++){
62 if(!vis[i][j]&&g[i][j]=='.')
63 dfs(i,j,++cc);
64 }
65 for(int i=sea+1;i<=cc;i++) {lake[++cnt].size=num[i];lake[cnt].id=i;}//printf("%d %d\n",sea,cc);
66 sort(lake+1,lake+1+cnt,cmp);
67 //for(int i=1;i<=cnt;i++) printf("lake %d %d\n",lake[i].id,lake[i].size);
68 int t=cnt-k,p=1;//printf("t %d\n",t);
69 for(int z=1;z<=t;z++){
70 int fin=0;
71 for(int i=1;i<=n;i++){
72 for(int j=1;j<=m;j++)
73 if(lake[p].id==vis[i][j]){fil(i,j,lake[p++].id);fin=1;break;}
74 if(fin) break;
75 }
76 }
77 printf("%d\n",ans);
78 for(int i=1;i<=n;i++){
79 for(int j=1;j<=m;j++) printf("%c",g[i][j]);
80 if(i!=n) putchar('\n');
81 }
82
83 }
F - Lakes in Berland(BFS)的更多相关文章
- 【29.70%】【codeforces 723D】Lakes in Berland
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- codeforce375div2-D. Lakes in Berland 搜索
Lakes in Berland 题意与解释:这道题就是求图中被围起来的点群,问最少去掉几个点,可以使得孤立的点群数目为K; 因为自己写的代码又长又had bugs. 我自己写的bfs,想着是先染色, ...
- CF723D. Lakes in Berland[DFS floodfill]
D. Lakes in Berland time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- codeforces 723D: Lakes in Berland
Description The map of Berland is a rectangle of the size n × m, which consists of cells of size 1 × ...
- CF723D 【Lakes in Berland】
题目链接 题解 CF723D [Lakes in Berland] 首先将边界的水用bfs处理掉 再将中间的每一个湖泊处理出来,存入一个结构体内,结构体里记录湖泊大小和开始点 将湖泊排序从小往大填满, ...
- cf723d Lakes in Berland
The map of Berland is a rectangle of the size n × m, which consists of cells of size 1 × 1. Each cel ...
- Codeforces Round #375 (Div. 2)——D. Lakes in Berland(DFS连通块)
D. Lakes in Berland time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #375 (Div. 2) D. Lakes in Berland dfs
D. Lakes in Berland time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- codeforces723 D. Lakes in Berland(并查集)
题目链接:codeforces723 D. Lakes in Berland 参考博客:http://www.cnblogs.com/Geek-xiyang/p/5930245.html #inclu ...
随机推荐
- 如何用css写一个带斜切角、有边框又有内外阴影的按钮呢?
如果有一天,UI设计师丢过来一张UI稿,上面有这样一个带有斜切角.有边框还有内外阴影的按钮,你会怎么实现呢?第一反应切图?可是按钮内容.大小都是可变的,那得切多少图啊~Canvas?SVG?No,no ...
- 分布式实时处理系统——C++高性能编程
[前言]基于通信基础,介绍Hurricane实时处理系统的工程实现,主要使用C++语言. 一.IPC.socket.异步I/O epoll 二.C++11 1.linux内存管理中使用RALL原则,C ...
- CF1491C Pekora and Trampoline 题解
题目链接 比赛时只想到了 \(\mathcal O(n^3)\) 的暴力做法,官方题解是 \(\mathcal O(n^2)\) ,并且是可以优化为 \(\mathcal O(n)\) 的(贪心+ ...
- Typora学习
Markdown学习总结 标题的使用格式 # 一阶标题 或者 ctrl + 1 ## 二阶标题 或者 ctrl + 2 ### 三阶标题 或者 ctrl + 3 #### 四阶标题 或者 ctrl + ...
- java关于字符串是否存
1, if('true'.equalsIgnoreCase(response.result as String)); 2, if (scvrsp.toLowerCase().contains(&q ...
- 记一次jstack命令定位问题
今天天气不错,但是赶上恶意加班心情就不爽,怀着不爽的心情干活,总能创造出更多的问题,这不,今天就自己挖了一个坑,自己跳进去了,好在上来了 经过是这样的,开始调试canal采集binlog时,由于添加了 ...
- 2019 南京网络赛 B super_log 【递归欧拉降幂】
一.题目 super_log 二.分析 公式很好推出来,就是$$a^{a^{a^{a^{...}}}}$$一共是$b$个$a$. 对于上式,由于指数太大,需要降幂,这里需要用到扩展欧拉定理: 用这个定 ...
- python--requests模块详解
GET请求 首先构造一个最简单的get请求,请求的链接为http://httpbin.org/get import requests 2 r = requests.get("http://h ...
- ch1_5_1统计最大最小元素的平均比较次数
public class ch1_5_1统计最大最小元素的平均比较次数 { public static void main(String[] args) { // TODO Auto-generate ...
- windows创建签名文件pfx
https://stackoverflow.com/questions/84847/how-do-i-create-a-self-signed-certificate-for-code-signing ...