POJ 2157

题意：迷宫里面有门和钥匙，门用大写字母表示，钥匙小写。开门必须拿到迷宫中所有这个门的钥匙。给出起始点和终点，求是否能有办法走到。

 

思路：floodfill + bfs。先找能拿到哪些钥匙，判断遇到哪些门，再判断哪些门可以被打开。如果没有走到终点，就继续找钥匙。如果没有门可开了，又没有走到终点，则输出"NO"

 

易错点：这个题的细节实在多。除了代码能力要求比较高以外，要判断很多特殊情况。其中一个就是，当迷宫中有一个门，但是没有对应的钥匙。一不留神，程序就会导致门被打开了。

 

统计：244k, 0ms, 3Y

 

#include <iostream>#include <string>#define F(i,a,b) for (int i=a;i<=b;i++)using namespace std;bool mk[21][21], meetdoor[6];int havek[6], keys[6], startx, starty, Q[402][2], move[4][2] = { {0,1}, {1,0}, {-1,0}, {0,-1} },M, N, door[6][2];string map[21];bool floodfill(){int doors = 0;memset(havek, 0, sizeof(havek) );memset(mk, 0, sizeof(mk) );int now = 1, last = 1;Q[1][0]= startx, Q[1][1]=starty;mk[startx][starty] =true;while (now<=last){F(i,0,3){int newx = Q[now][0]+move[i][0], newy=Q[now][1]+move[i][1];if (newx >=0 && newx<M && newy>=0 && newy<N){if (mk[newx][newy] || map[newx][newy] =='X')continue;if (map[newx][newy] >=65 && map[newx][newy]<=69 ){if (!meetdoor[ map[newx][newy] - 65 ] ){doors++;meetdoor[ map[newx][newy] - 65 ] =true;}continue;}mk[newx][newy] = true;Q[++last][0] = newx;Q[last][1] = newy;char ch = map[newx][newy];//keyif (ch>=97 && ch<=101)havek[ch-97]++;}}now++;}F(i,0,4){if (meetdoor[i] && havek[i] == keys[i] && keys[i] !=0 )map[ door[i][0] ].replace( door[i][1], 1, 1, '.');}return (doors!=0);}bool bfs(){memset(mk, 0 , sizeof(mk ) );int now = 1, last = 1;Q[1][0] = startx, Q[1][1] = starty;mk[startx][starty] = true;while (now <= last){F(i,0,3){int newx = Q[now][0]+move[i][0], newy=Q[now][1]+move[i][1];if (newx >=0 && newx<M && newy>=0 && newy<N){char ch = map[newx][newy];if (!mk[newx][newy] && ch !='X'){if (ch=='.' || (ch>=97 && ch<=101)){Q[++last][0] = newx;Q[last][1] = newy;mk[newx][newy] = true;}if (ch =='G')return true;}}}now ++;}return false;}int main(){while ( cin >> M >> N && ( M!=0 && N!=0 ) ){memset(keys, 0, sizeof(keys) );memset(havek, 0, sizeof(havek) );memset(meetdoor, 0, sizeof(meetdoor) );F(i,0,M-1){cin >> map[i];F(j,0,N-1){if (map[i][j] >=97 && map[i][j] <=101)keys[ map[i][j] - 97 ]++;if (map[i][j]>=65 && map[i][j]<=69)door[ map[i][j]-65 ][0] =i, door[ map[i][j] -65 ][1] = j;if (map[i][j]=='S')startx = i, starty = j;}}while (1){bool finddoor = floodfill();if ( bfs() ){cout << "YES/n";break;}if (!finddoor){cout << "NO/n";break;}}}return 0;}