hdu 5025 Saving Tang Monk 状态压缩dp+广搜

作者：jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4092939.html

题目链接：hdu 5025 Saving Tang Monk 状态压缩dp+广搜

使用dp[x][y][key][s]来记录孙悟空的坐标(x,y)、当前获取到的钥匙key和打死的蛇s。由于钥匙具有先后顺序，因此在钥匙维度中只需开辟大小为10的长度来保存当前获取的最大编号的钥匙即可。蛇没有先后顺序，其中s的二进制的第i位等于1表示打死了该蛇，否则表示没打死。然后进行广度优先搜索即可，需要注意的是即使目前没有拿到第k-1把钥匙，也可以经过房间k或唐僧，只不过无法获取钥匙k和救唐僧而已。

代码如下：

 #include <cstdio>
 #include <cstdlib>
 #include <iostream>
 #include <cstring>
 #include  <queue>
 #include  <limits.h>
 #define     MAXN 101
 #define     MAXM 10
 using namespace std;
 int dir[][]={{, }, {, -}, {-, }, {, }};
 char map[MAXN][MAXN];
 int bin[] = {, , , , , , , , , , , , };
 int dp[MAXN][MAXN][][];
 int n, m;
 class state
 {
     public:
         int x, y, step, key, s;
 };
 state start, ed;
 void solve()
 {
     memset(dp, -, sizeof(dp));
     queue<state> qu;
     qu.push(start);
     int res = INT_MAX;
     while( qu.size() >  )
     {
         state cur = qu.front();
         qu.pop();
         for( int i =  ; i <  ; i++ )
         {
             if( cur.x == ed.x && cur.y == ed.y && cur.key== m )
             {
                 res = min(res, cur.step);
                 continue;
             }
             state next;
             next.x = cur.x+dir[i][];
             next.y = cur.y+dir[i][];
             next.key = cur.key;
             next.s = cur.s;
             next.step = cur.step;
             if( next.x <  || next.x >= n || next.y <  || next.y >= n )//出界
             {
                 continue;
             }
             char tmp = map[next.x][next.y];
             if( tmp  == '#' )//陷阱
             {
                 continue;
             }
             if( tmp >= '' && tmp <= '' && cur.key >= tmp-''- )//有钥匙
             {
                 next.step = cur.step+;
                 next.key = max(cur.key, tmp - '');
                 next.s = cur.s;
             }
             else if( tmp >= 'A' && tmp <= 'F')//蛇
             {
                 if( cur.s/bin[tmp-'A']% ==  )
                 {
                     next.step = cur.step + ;
                 }
                 else
                 {
                     next.step = cur.step + ;
                 }
                 next.key = cur.key;
                 next.s = cur.s | bin[tmp-'A'];
             }
             else
             {
                 next.key = cur.key;
                 next.step = cur.step+;
                 next.s = cur.s;
             }
             int dptmp = dp[next.x][next.y][next.key][next.s];
             if( dptmp <  )
             {
                 dp[next.x][next.y][next.key][next.s] = next.step;
                 qu.push(next);
             }
             else if(dptmp > next.step)
             {
                 dp[next.x][next.y][next.key][next.s] = next.step;
                 qu.push(next);
             }
         }
     }
     if( res == INT_MAX )
     {
         printf("impossible\n");
         return;
     }
     printf("%d\n", res);
 }
 int main(int argc, char *argv[])
 {
     char tmp[MAXN+];
     while()
     {
         scanf("%d%d", &n, &m);
         if( n ==  && m ==  ) return ;
         int sn = ;
         for( int i =  ; i < n ; i++ )
         {
             scanf("%s", tmp);
             for( int j =  ; j < n ; j++ )
             {
                 if( tmp[j] == 'K' )
                 {
                     start.x = i;
                     start.y = j;
                     start.key = ;
                     start.step = ;
                     start.s = ;
                 }
                 else if( tmp[j] == 'T' )
                 {
                     ed.x = i;
                     ed.y = j;
                 }
                 else if( tmp[j] == 'S' )
                 {
                     map[i][j] = 'A'+sn;
                     sn++;
                     continue;
                 }
                 map[i][j] = tmp[j];
             }
         }
         solve();
     }
 }