poj-1164 The Castle

最新推荐文章于 2021-05-06 15:36:49 发布

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最新推荐文章于 2021-05-06 15:36:49 发布

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原文链接：http://www.cnblogs.com/zmin/p/7340151.html

本文介绍了一种通过深度优先搜索算法来计算迷宫中房间数量及最大房间面积的方法。利用二进制位运算判断墙壁方向，实现对地图的遍历，并统计房间数量和最大房间覆盖的模块数。

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     1   2   3   4   5   6   7  

   #############################

 1 #   |   #   |   #   |   |   #

   #####---#####---#---#####---#

 2 #   #   |   #   #   #   #   #

   #---#####---#####---#####---#

 3 #   |   |   #   #   #   #   #

   #---#########---#####---#---#

 4 #   #   |   |   |   |   #   #

   #############################

(Figure 1)


#  = Wall   

|  = No wall

-  = No wall

Figure 1 shows the map of a castle.Write a program that calculates
1. how many rooms the castle has
2. how big the largest room is
The castle is divided into m * n (m<=50, n<=50) square modules. Each such module can have between zero and four walls.

Input

Your program is to read from standard input. The first line contains the number of modules in the north-south direction and the number of modules in the east-west direction. In the following lines each module is described by a number (0 <= p <= 15). This number is the sum of: 1 (= wall to the west), 2 (= wall to the north), 4 (= wall to the east), 8 (= wall to the south). Inner walls are defined twice; a wall to the south in module 1,1 is also indicated as a wall to the north in module 2,1. The castle always has at least two rooms.

Output

Your program is to write to standard output: First the number of rooms, then the area of the largest room (counted in modules).

Sample Input

4
7
11 6 11 6 3 10 6
7 9 6 13 5 15 5
1 10 12 7 13 7 5
13 11 10 8 10 12 13

Sample Output

5
9

这题的题面很长，大致意思是每个坐标有东南西北四个方位，每个方位都可能有墙。1，2，4，8分别是西，北，东，南
四个方位有墙，输入的每位数字是墙的标号之和，比如11是1+2+8，就是西，北，南有墙，要求算的是有多少个连通块
和最大连通块连了几个坐标。
这四个数字很特殊，没错，就是用二进制的位运算判断。
附代码：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int M = 55;
int map[M][M];
int vis[M][M];
int n,m,maxx=0,cnt=0,res=0;
void dfs(int x,int y){
    
    if((y-1)>=1&&!(map[x][y]&1)&&!vis[x][y-1]){
     //   cout<<1<<endl;
        vis[x][y-1]=1;
        cnt++;
        dfs(x,y-1);
    }
    if((x-1)>=1&&!(map[x][y]&2)&&!vis[x-1][y]){
     //   cout<<2<<endl;
        vis[x-1][y]=1;
        cnt++;
        dfs(x-1,y);
    }
    if((y+1)<=m&&!(map[x][y]&4)&&!vis[x][y+1]){
     //   cout<<3<<endl;
        vis[x][y+1]=1;
        cnt++;
        dfs(x,y+1);
    }
    if((x+1)<=n&&!(map[x][y]&8)&&!vis[x+1][y]){
     //   cout<<4<<endl;
        vis[x+1][y]=1;
        cnt++;
        dfs(x+1,y);
    }
}
int main(){
    while(cin>>n>>m){
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                cin>>map[i][j];
            }
        }
        maxx=0;res=0;cnt=1;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(!vis[i][j]){
                    if(map[i][j]!=15)
                    {
            //        cout<<i<<" "<<j<<endl;
                    vis[i][j]=1;
                    dfs(i,j);
                    
                    maxx=max(cnt,maxx);
                  //  cout<<cnt<<" maxx"<<endl;
                    cnt=1;
                    }
                    res++;
                }
            }
        }
        cout<<res<<endl<<maxx<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/zmin/p/7340151.html