E The Binding of Isaac

最新推荐文章于 2018-09-09 15:22:36 发布

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本文介绍了一款迷宫寻宝游戏的算法实现，玩家需要帮助角色Isaac找到隐藏的超级秘密房间。通过分析地图布局，算法计算出可能藏匿秘密房间的位置数量。

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

题解: 找到 ‘.' 周围只有一个 ' # ' 的 '.' 的个数；注意四周都认为有 '.' ;

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int  n, m; 
char s[150][150];

bool f(int i, int j){
	int amount = 0;
	if(s[i][j] == '.'){
		if(i - 1 >= 0 && s[i - 1][j] == '#') amount++;
		if(i + 1 <= n + 1 && s[i + 1][j] == '#') amount++;
		if(j - 1 >= 0 && s[i][j - 1] == '#') amount++;
		if(j + 1 <= m + 1 && s[i][j + 1] == '#') amount++;
		if(amount == 1)
			return true;
		else 
			return false;
	}
	return false;
}

int main(){
	int T, i, j;
	cin >> T;
	while (T--){
		cin >> n >> m;
		for(i = 1; i <= n; ++i){
		getchar();
			for(j = 1; j <= m; ++j){
				scanf("%c", &s[i][j]);
			}
			
		}
		for(i = 0; i <= m + 1; i++){
			s[0][i] = s[n + 1][i] = '.';
		}
		for(j = 0; j <= n + 1; ++j){
			s[j][0] = s[j][m + 1] = '.';
		}
		int result = 0;
		for(i = 0; i <= n + 1; ++i){
			for(j = 0; j <= m + 1; ++j){
				if(f(i, j))
					++result;
					
			}
		}
		cout << result << endl;
	} 
	return 0;
}