主要就是开个 三维vis数组 第三维 表示在<x,y>坐标下的 钥匙有哪些
还有要注意的话 就是wall数组 是有3种不同的情况 -1 直接行走 0 墙 >=1 大门 需要钥匙
1 #include <iostream> 2 #include <cstring> 3 #include <queue> 4 using namespace std; 5 6 int n , m; 7 const int size = 55; 8 int wall[size][size][size][size]; 9 int dir[4][2] = {1,0,-1,0,0,1,0,-1}; 10 int key[size][size]; 11 bool vis[size][size][1<<12]; 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( vis , false , sizeof(vis) ); 23 memset( wall , -1 , sizeof(wall) ); 24 memset( key , 0 , sizeof(key) ); 25 while( !q.empty() ) 26 q.pop(); 27 } 28 29 int bfs( int x1 , int y1 , int x2 , int y2 ) 30 { 31 node now; 32 q.push( node(x1,y1,key[x1][y1],0) ); 33 vis[x1][y1][ key[x1][y1] ] = true; 34 while( !q.empty() ) 35 { 36 now = q.front(); 37 q.pop(); 38 if( now.x==x2 && now.y==y2 ) 39 return now.T; 40 for( int i = 0 ; i<4 ; i++ ) 41 { 42 int xx = now.x + dir[i][0]; 43 int yy = now.y + dir[i][1]; 44 if( xx>=1 && xx<=n && yy>=1 && yy<=m ) 45 { 46 int door = wall[xx][yy][now.x][now.y]; 47 int kk = now.key | key[xx][yy]; 48 if( !vis[xx][yy][kk] && ( door==-1 || ( door>=1 && ( (1<<door)&now.key )!=0 ) ) ) 49 { 50 vis[xx][yy][kk] = true; 51 q.push( node(xx,yy,kk,now.T+1) ); 52 } 53 } 54 } 55 } 56 return -1; 57 } 58 59 int main() 60 { 61 int p , k , s , g , q , x1 , x2 , y1 , y2 , ans; 62 while( cin >> n >> m >> p ) 63 { 64 cin >> k; 65 init(); 66 while( k-- ) 67 { 68 cin >> x1 >> y1 >> x2 >> y2 >> g; 69 wall[x1][y1][x2][y2] = wall[x2][y2][x1][y1] = g; 70 } 71 cin >> s; 72 while( s-- ) 73 { 74 cin >> x1 >> y1 >> q; 75 key[x1][y1] |= (1<<q); 76 } 77 ans = bfs( 1 , 1 , n , m ); 78 cout << ans << endl; 79 } 80 return 0; 81 }
再去做下 上次的 二分图 =_=