第七届山东省ACM The Binding of Isaac（搜索OR暴力。。）

The Binding of Isaac

Time Limit: 2000MS Memory Limit: 65536KB

Problem Description

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.

Example Input

2
5 3
..#
.##
##.
.##
##.
1 1
#

Example Output

8
4

Hint

Author

“浪潮杯”山东省第七届ACM大学生程序设计竞赛

题意：给你一个地图，判断super-secret room 的数量，规定超秘房间的四周只有一个common room.

直接暴力搜索就好了，没含量，不过注意输入输出的小细节！！！

#include <bits/stdc++.h>
using namespace std;
#define MAX 110
char mmap[MAX][MAX];
int dirX[]= {-1,1,0,0};
int dirY[]= {0,0,-1,1};
int main()
{
    int T,n,m,sum;
    char ch[MAX];
    scanf("%d",&T);

    while(T--)
    {
        sum=0;
        scanf("%d%d",&m,&n);
        for(int i=0; i<m+2; i++)  ///整张图初始化
            for(int j=0; j<n+2; j++)
                mmap[i][j]='.';
/*
        getchar();   ///用输入字符的方式
        for(int i=1; i<=m; i++){
            for(int j=1;j<=n;j++)
              scanf("%c",&mmap[i][j]);
            getchar();
        }*/
         for(int i=1; i<=m; i++)
            scanf("%s",mmap[i]+1);
         for(int i=0;i<m+2;i++)   ///注意最后的
            mmap[i][n+1]='.';
         
        for(int i=0; i<m+2; i++)
            for(int j=0; j<n+2; j++)
                if(mmap[i][j] == '.')
                {
                    int tmp=0;
                    for(int k=0; k<4; k++)  ///判断四个方向
                        if( (i+dirX[k]>=0&&i+dirX[k]<=m+1) && (j+dirY[k]>=0&&j+dirY[k]<=n+1) && mmap[i+dirX[k]][j+dirY[k]]=='#')
                            tmp++;
                    if(tmp==1)
                }
        printf("%d
",sum);

    }
    return 0;
}