hdu--1885--状压bfs

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 }
View Code

 

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

posted @ 2014-11-16 15:55  radical  阅读(170)  评论(0)  编辑  收藏  举报