山东省第七届ACM省赛------The Binding of Isaac

The Binding of Isaac

Time Limit: 2000MS Memory limit: 65536K

题目描述

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.

输入

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.

输出

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

示例输入

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

示例输出

8
4

来源

“浪潮杯”山东省第七届ACM大学生程序设计竞赛

题意

题目虽然看起来很长，但是只要读懂它就非常简单啦～
说的是给你一个地图，’.‘代表空地，‘#’代表门，求在地图中某个非‘#’的点所相邻的上下左右总共有一个‘#’的位置有多少个。
我们可以将整个地图保持在一个字符二维数组中，然后在输入的时候向右向下分别偏移一个单位，因为地图外的一周也要检测啦～
然后枚举所有不是‘#’的点便可以了，反正不会超时，最后输出结果～

AC代码：

#include"stdio.h"
#include"string.h"
#include<iostream>
using namespace std;
char a[105][105];
int main()
{
int n;
cin>>n;
while(n--)
{
int c,b;
cin>>c>>b;
getchar();
memset(a,0,sizeof(a));
for(int i=1; i<=c; i++)
gets(a[i]+1);
int s=0;
for(int i=0; i<=c+1; i++)
for(int j=0; j<=b+1; j++)
if(a[i][j]!='#')
{
int d=0;
if(i>0&&a[i-1][j]=='#')d++;
if(a[i+1][j]=='#')d++;
if(j>0&&a[i][j-1]=='#')d++;
if(a[i][j+1]=='#')d++;
if(d==1)s++;
}
printf("%d\n",s);
}
return 0;
}