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最新推荐文章于 2023-06-21 21:33:19 发布

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博客围绕游戏《以撒的结合》展开，介绍游戏里Isaac被困迷宫，要找超级密室，该密室在仅与一个普通房间相邻的空位。给出输入输出要求及示例，还提及计算方法，即遍历方格，边上遇#加一，里面判定上下左右#数量。

The Binding of Isaac
Ok, now I will introduce this game to you…

Isaac is trapped in a maze which has many common rooms…

Like this…There are 9 common rooms on the map.
And there is only one super-secret room. We can’t see it on the map. The super-secret room always has many special items in it. Isaac wants to find it but he doesn’t know where it is.Bob

tells him that the super-secret room is located in an empty place which is adjacent to only one common rooms.

Two rooms are called adjacent only if they share an edge. But there will be many possible places.
Now Isaac wants you to help him to find how many places may be the super-secret room.

Input

Multiple test cases. The first line contains an integer T (T<=3000), indicating the number of test case.

Each test case begins with a line containing two integers N and M (N<=100, M<=100) indicating the number

of rows and columns. N lines follow, “#” represent a common room. “.” represent an empty place.Common rooms

maybe not connect. Don’t worry, Isaac can teleport.

Output

One line per case. The number of places which may be the super-secret room.

Sample

Input
Copy2
5 3
…#
.##
##.
.##
##.
1 1

Output

Copy8
4

遍历每个小方格；边上的单独遍历，只要是#就加一；里面的要判定上下左右有多少个#。

#include <bits/stdc++.h>
using namespace std;
int t,n,m,i,j,a[120][120];
int fun(int i,int j)
{
    int k=0;
    if(a[i-1][j]==1)
        k++;
    if(a[i][j-1]==1)
        k++;
    if(a[i+1][j]==1)
        k++;
    if(a[i][j+1]==1)
        k++;
    if(k==1)
        return 1;
    else
        return 0;
}
int main()
{
    char c;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        scanf("%d%d",&n,&m);
        getchar();
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                scanf("%c",&c);
                if(c=='.')
                    a[i][j]=0;
                else if(c=='#')
                    a[i][j]=1;
            }
            getchar();
        }
        int cnt=0;
        for(i=1; i<=n; i++)
        {
            if(a[i][1])
                cnt++;
            if(a[i][m])
                cnt++;
        }
        for(j=1; j<=m; j++)
        {
            if(a[1][j])
                cnt++;
            if(a[n][j])
                cnt++;
        }
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
            {
                if(a[i][j]==0)
                {
                    if(fun(i,j))
                        cnt++;
                }
            }
        }
        cout<<cnt<<'\n';
    }
    return 0;
}