这里 我花了将近10分钟找错 才发现是错在 运算符的优先级上面 =_=
1 if( mp[xx][yy]=='B' && ( (now.key&2)==0) )
位运算的优先级 太低了 ....
每次多开一维来表示该点的 钥匙数 很多类似~~
1 #include <iostream> 2 #include <cstring> 3 #include <queue> 4 using namespace std; 5 6 int n , m; 7 const int size = 110; 8 int dir[4][2] = {1,0,-1,0,0,1,0,-1}; 9 char mp[size][size]; 10 int kk[size][size]; 11 bool vis[size][size][1<<6]; 12 struct node 13 { 14 int x , y , key , T; 15 node(){}; 16 node( int u , int v , int w , int p ):x(u),y(v),key(w),T(p){}; 17 }; 18 queue<node> q; 19 20 void init( ) 21 { 22 memset( kk , 0 , sizeof(kk) ); 23 memset( vis , false, sizeof(vis) ); 24 while( !q.empty() ) 25 q.pop(); 26 } 27 28 int bfs( int stx , int sty ) 29 { 30 q.push( node( stx , sty , kk[stx][sty] , 0 ) ); 31 vis[stx][sty][ kk[stx][sty] ] = true; 32 while( !q.empty() ) 33 { 34 node now = q.front(); 35 q.pop(); 36 //cout << "x: " << now.x << " y: " << now.y << " T: " << now.T << endl; 37 if( mp[ now.x ][ now.y ] == 'X' ) 38 return now.T; 39 for( int i = 0 ; i<4 ; i++ ) 40 { 41 int xx = now.x + dir[i][0]; 42 int yy = now.y + dir[i][1]; 43 int KEY = now.key | kk[ xx ][ yy ]; 44 if( xx>=0 && xx<n && yy>=0 && yy<m && mp[xx][yy]!='#' && !vis[xx][yy][KEY] ) 45 { 46 if( mp[xx][yy]=='B' && ( (now.key&2)==0) ) 47 continue; 48 else if( mp[xx][yy]=='Y' && ( (now.key&4)==0) ) 49 continue; 50 else if( mp[xx][yy]=='R' && ( (now.key&8)==0) ) 51 continue; 52 else if( mp[xx][yy]=='G' && ( (now.key&16)==0) ) 53 continue; 54 q.push( node(xx,yy,KEY,now.T+1) ); 55 vis[xx][yy][KEY] = true; 56 } 57 } 58 } 59 return -1; 60 } 61 62 int main() 63 { 64 cin.sync_with_stdio(false); 65 int ans , stx , sty; 66 while( cin >> n >> m &&(n||m) ) 67 { 68 init(); 69 for( int i = 0 ; i<n ; i++ ) 70 { 71 cin >> mp[i]; 72 } 73 for( int i = 0 ; i<n ; i++ ) 74 { 75 for( int j = 0 ; j<m ; j++ ) 76 { 77 if( mp[i][j]=='*' ) 78 { 79 stx = i; 80 sty = j; 81 } 82 else if( mp[i][j]=='b' ) 83 kk[i][j] |= (1<<1); 84 else if( mp[i][j]=='y' ) 85 kk[i][j] |= (1<<2); 86 else if( mp[i][j]=='r' ) 87 kk[i][j] |= (1<<3); 88 else if( mp[i][j]=='g' ) 89 kk[i][j] |= (1<<4); 90 } 91 } 92 ans = bfs( stx , sty ); 93 if( ans==-1 ) 94 cout << "The poor student is trapped!" << endl; 95 else 96 cout << "Escape possible in " << ans << " steps." << endl; 97 } 98 return 0; 99 }
today:
你必须先要失去什么你视之为珍贵的
才能得到一些更为有价值即使现在的你看来一文不值的