Lakes in Berland

題意與解釋：這道題就是求圖中被圍起來的點群，問最少去掉幾個點，可以使得孤立的點群數目為K;

　　　　　　因為自己寫的代碼又長又had bugs。

　　　　　　我自己寫的bfs，想著是先染色，後期在考慮這個顏色要不要留。

#include <iostream>
#include 
 
<cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include 
 
<bitset>
#include <cctype>
using namespace std;

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue
// #pragma comment(linker, "/STACK:10240000000,10240000000")
 
//擴棧，要用c++交，用g++交並沒有什麽卵用。。
typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;

#define fi first
#define se second

#define OKC ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用來壓行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)


const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf  = 0x3f3f3f3f;
template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
///*-----------------show time----------------*/
            const int maxn = 55;
            int mp[maxn][maxn],col[maxn][maxn],sp[maxn][maxn];
            int book[maxn][maxn];
            string g[maxn];
            // int a[3000];
            int nxt[5][5] {
                {1,0},
                {0,1},
                {-1,0},
                {0,-1}
            };
            int n,m,k;
            queue<pii>q;
            int bfs(int x,int y,int tug){
                    int mx = 1;
                    while(!q.empty())q.pop();
                    q.push(make_pair(x,y));
                    col[x][y] = tug;
                   // sp[x][y] = 1;
                    book[x][y] = 1;
                    while(!q.empty()){
                        int tx = q.front().fi;
                        int ty = q.front().se;
                        // cout<<tug<<"###"<<tx<<" "<<ty<<endl;
                        q.pop();
                        for(int i=0; i<=3; i++){
                            int nx = tx + nxt[i][0];
                            int ny = ty + nxt[i][1];  //這裏ty 寫成tx
                            if(nx < 0 || nx >= n || ny < 0 || ny >= m)continue;
                            if(mp[nx][ny] == 0)continue;
                            if(mp[nx][ny] == 1 && book[nx][ny] != 1){

                                col[nx][ny] = tug;
                            //    sp[nx][ny] = sp[tx][ty] + 1;
                            
                                book[nx][ny] = 1;
                                mx++;
                                if(nx == 0||nx == n-1||ny == 0||ny == m-1){
                                    mx = inf;                  //切莫不要直接return！
                                }
                                q.push(make_pair(nx,ny));
                            }
                        }
                        // debug(q.size());
                    }
                    return mx;
            }

            struct node{
                    int val;
                    int se;
            }a[3000];

            bool cmp(node a,node b){
                return a.val < b.val;
            }

            int shak[3000];
int main(){
            
            cin>>n>>m>>k;
            for(int i=0; i<n; i++){
                cin>>g[i];
                for(int j=0; j<m; j++){
                    if(g[i][j]==‘*‘) mp[i][j] = 0;
                    else mp[i][j] = 1;
                }
            }

            int tot = 0,  cc = 0;
            for(int i = 1; i<n-1; i++){
                for(int j = 1; j<m-1 ;j++){
                    if(book[i][j]!=1 && mp[i][j]){
                        cc++;
                        int d = bfs(i,j,cc);
                        if(d<inf){
                            tot++;
                            a[tot].val = d;
                            a[tot].se = cc;
                        }
                    }
                }
            }
            
            sort(a+1,a+1+tot,cmp);
            int sa = tot - k;
            int ans = 0;
            for(int i=1; i<=sa; i++){
                if(a[i].val < inf){
                    ans += a[i].val;
                    shak[a[i].se] = 1;
                }
            }
            
            printf("%d\n",ans);
            // debug(col[1][2]);
            for(int i=0; i<n; i++){
                for(int j=0;j<m; j++){
                    if(mp[i][j]==1)
                    {
                        if(shak[col[i][j]] == 1)
                            cout<<"*";
                        else cout<<".";
                    }
                    else cout<<"*";
                }
                cout<<endl;
            }
            return 0;
}

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