CodeForces 723D Lakes in Berland (dfs搜索)

题意：给定一个n*m的矩阵，*表示陆地， . 表示水，一些连通的水且不在边界表示湖，让你填最少的陆地使得图中湖剩下恰好为k。

析：很简单的一个搜索题，搜两次，第一次把每个湖的位置和连通块的数量记下来，第二次去填陆地，选少的进行填。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e3 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s[55][55];
struct Node{
    int x, y, cnt;
    Node(int xx, int yy, int c) : x(xx), y(yy), cnt(c) { }
    bool operator < (const Node &p) const{
        return cnt < p.cnt;
    }
};
bool vis[55][55];
int ans, cnt;

void dfs(int r, int c){
    ans = Max(ans, cnt);
    for(int i = 0; i < 4; ++i){
        int x = r + dr[i];
        int y = c + dc[i];
        if(x < 0 || x >= n || y < 0 || y >= m){ ans = INF;  continue; }
        if(vis[x][y] || s[x][y] == '*')  continue;
        vis[x][y] = true;
        ++cnt;
        dfs(x, y);
    }
}

void dfs1(int r, int c){
    for(int i = 0; i < 4; ++i){
        int x = r + dr[i];
        int y = c + dc[i];
        if(!is_in(x, y) || s[x][y] == '*') continue;
        s[x][y] = '*';
        dfs1(x, y);
    }
}

int main(){
    int k;
    while(scanf("%d %d %d", &n, &m, &k) == 3){
        for(int i = 0; i < n; ++i)  scanf("%s", s+i);

        vector<Node> v;
        memset(vis, false, sizeof vis);
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < m; ++j){
                if(s[i][j] == '.' && !vis[i][j]){
                    ans = 0;
                    cnt = 1;
                    vis[i][j] = true;
                    dfs(i, j);
                    if(ans != INF)  v.push_back(Node(i, j, ans));
                }
            }
        }

        sort(v.begin(), v.end());
        ans = 0;
        for(int i = 0; i+k < v.size(); ++i){
            s[v[i].x][v[i].y] = '*';
            dfs1(v[i].x, v[i].y);
            ans += v[i].cnt;
        }

        printf("%d\n", ans);
        for(int i = 0; i < n; ++i)
            puts(s[i]);
    }
    return 0;
}