Description

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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.

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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.

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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

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

Sample Output

8 4 /// /// _ooOoo_ /// o8888888o /// 88" . "88 /// (| -_- |) /// O\ = /O /// ____/`---'\____ /// .' \\| |// `. /// / \\||| : |||// \ /// / _||||| -:- |||||- \ /// | | \\\ - /// | | /// | \_| ''\---/'' | | /// \ .-\__ `-` ___/-. / /// ___`. .' /--.--\ `. . __ /// ."" '< `.___\_<|>_/___.' >'"". /// | | : `- \`.;`\ _ /`;.`/ - ` : | | /// \ \ `-. \_ __\ /__ _/ .-` / / /// ======`-.____`-.___\_____/___.-`____.-'====== /// `=---=' /// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ /// Buddha Bless, No Bug ! /// #include <cstdio> #include <iostream> #include <string> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <vector> #include <map> using namespace std; #define ll long long const int mod = 1e9+7;int t, n, m, sum, mov[4][2] = {0,1, 0,-1, 1,0, -1,0}; char fig[200+8][200+8];int main() {for(scanf("%d", &t); t--; ){scanf("%d%d", &n, &m);getchar();sum = 0;for(int i = 0; i <= n+1; i++)for(int j = 0; j <= m+1; j++)fig[i][j] = '.';for(int i = 1; i <= n; i++){scanf("%s", fig[i]+1);getchar();}for(int i = 0; i <= n+1; i++)///后來總是無法消去回車鍵，所以要另外補充 '.'fig[i][m+1] = '.';for(int i = 0; i <= n+1; i++){for(int j = 0; j <= m+1; j++){int num = 0;if(fig[i][j] == '.'){for(int k = 0; k<4; k++){int buffer_x, buffer_y;buffer_x = i+mov[k][0];buffer_y = j+mov[k][1];if(buffer_x >= 0 && buffer_x <= n+1 && buffer_y >= 0 && buffer_y <= m+1 && fig[buffer_x][buffer_y] == '#')num++;}if(num == 1)sum++;}}}printf("%d\n", sum);}return 0; }

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轉載于:https://www.cnblogs.com/RootVount/p/11237911.html

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