Codeforces Round #375 (Div. 2) D 连通块

来源：互联网发布：初榨椰子油知乎编辑：程序博客网时间：2024/05/16 11:36

题目传送门：http://codeforces.com/contest/723/problem/D

题意：给一张N*M的图，图被ocean包围，'*'代表land，‘.’代表cell，相邻的所有cell组成lake，如果有cell靠近海，则其所有相连的cell不能组成lake，给一个K，问最少进行多少次操作将cell变成land，才能让图只有K个lake（k一定小于等于现有lake）

思路：就是连通图，比赛时不如先做这一道。但赛后做还是wa了n发，思路没错就是有些细节马虎了，宛如一个zz = =

代码如下：

#include <iostream>#include <algorithm>#include <cstring>#include <stdio.h>#include <string>#include <cmath>#include <queue>#include <set>#include <map>using namespace std;#define   lson          l,m,rt<<1#define   rson          m+1,r,rt<<1|1#define   ll            long long#define   ull           unsigned long long#define   mem(n,v)      memset(n,v,sizeof(n))#define   MAX           55#define   MAXN          3000#define   PI            3.1415926#define   E             2.718281828459#define   opnin         freopen("text.in.txt","r",stdin)#define   opnout        freopen("text.out.txt","w",stdout)#define   clsin         fclose(stdin)#define   clsout        fclose(stdout)#define   haha1         cout << "haha1"<< endl#define   haha2         cout << "haha2"<< endl#define   haha3         cout << "haha3"<< endlconst int    INF    =   0x3f3f3f3f;const ll     INFF   =   0x3f3f3f3f3f3f3f3f;const double pi     =   3.141592653589793;const double inf    =   1e18;const double eps    =   1e-8;const ll     mod    =   1e9+7;const ull    mx     =   133333331;int Map[MAX][MAX];int vis[MAX][MAX];int n,m,k;struct llake{    int x,y,Size;    llake(int a = 0,int b = 0,int S = 0) : x(a),y(b),Size(S) {}    friend bool operator < (llake a,llake b)    {        return a.Size < b.Size;    }}lake[MAXN];int flag = false;int dfs(int x,int y){    int Size = 1;    vis[x][y] = 1;    if(x == 1 || x == n || y == 1 || y == m){        flag = true;    }    if(x - 1 >= 1 && Map[x-1][y] == '.' && !vis[x-1][y] ) Size += dfs(x-1,y);    if(x + 1 <= n && Map[x+1][y] == '.' && !vis[x+1][y] ) Size += dfs(x+1,y);    if(y - 1 >= 1 && Map[x][y-1] == '.' && !vis[x][y-1] ) Size += dfs(x,y-1);    if(y + 1 <= m && Map[x][y+1] == '.' && !vis[x][y+1] ) Size += dfs(x,y+1);    return Size;}void dfs1(int x,int y){    Map[x][y] = '*';    if(x - 1 > 1 && Map[x-1][y] == '.')  dfs1(x-1,y);    if(x + 1 < n && Map[x+1][y] == '.')  dfs1(x+1,y);    if(y - 1 > 1 && Map[x][y-1] == '.')  dfs1(x,y-1);    if(y + 1 < m && Map[x][y+1] == '.')  dfs1(x,y+1);}int main(){    cin >> n >> m >> k;    mem(Map,0);    mem(vis,0);    mem(lake,0);    int num = 1;    getchar();    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            scanf("%c",&Map[i][j]);        }        getchar();    }    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            if(Map[i][j] == '.' && !vis[i][j]){                flag = false;                int temp = dfs(i,j);                if(!flag && temp > 0){                    lake[num].x = i;                    lake[num].y = j;                    lake[num++].Size = temp;                }            }        }    }    int cnt = 0;    int num_temp = num;    sort(lake+1,lake+num+1);    for(int i=1;i<=num;i++){        if(num_temp == k)  break;        cnt += lake[i].Size;        num_temp--;        dfs1(lake[i].x,lake[i].y);    }    cout << cnt << endl;    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            printf("%c",Map[i][j]);        }        printf("\n");    }    return 0;}

0 0