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：

 

  1 #include <cstdio>

  2 #include <cstdlib>

  3 #include <iostream>

  4 #include <cstring>

  5 #include  <queue>

  6 #include  <limits.h>

  7 #define     MAXN 101

  8 #define     MAXM 10

  9 using namespace std;

 10 int dir[4][2]={{0, 1}, {0, -1}, {-1, 0}, {1, 0}};

 11 char map[MAXN][MAXN];

 12 int bin[] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096};

 13 int dp[MAXN][MAXN][10][35];

 14 int n, m;

 15 class state

 16 {

 17     public:

 18         int x, y, step, key, s;

 19 };

 20 state start, ed;

 21 void solve()

 22 {

 23     memset(dp, -1, sizeof(dp));

 24     queue<state> qu; 

 25     qu.push(start);

 26     int res = INT_MAX;

 27     while( qu.size() > 0 )

 28     {

 29         state cur = qu.front();

 30         qu.pop();

 31         for( int i = 0 ; i < 4 ; i++ )

 32         {

 33             if( cur.x == ed.x && cur.y == ed.y && cur.key== m )

 34             {

 35                 res = min(res, cur.step);

 36                 continue;

 37             }

 38             state next;

 39             next.x = cur.x+dir[i][0];

 40             next.y = cur.y+dir[i][1];

 41             next.key = cur.key;

 42             next.s = cur.s;

 43             next.step = cur.step;

 44             if( next.x < 0 || next.x >= n || next.y < 0 || next.y >= n )//出界

 45             {

 46                 continue;

 47             }

 48             char tmp = map[next.x][next.y];

 49             if( tmp  == '#' )//陷阱

 50             {

 51                 continue;

 52             }

 53             if( tmp >= '1' && tmp <= '9' && cur.key >= tmp-'0'-1 )//有钥匙

 54             {

 55                 next.step = cur.step+1;

 56                 next.key = max(cur.key, tmp - '0');

 57                 next.s = cur.s;

 58             }

 59             else if( tmp >= 'A' && tmp <= 'F')//蛇

 60             {

 61                 if( cur.s/bin[tmp-'A']%2 == 1 )

 62                 {

 63                     next.step = cur.step + 1;    

 64                 }

 65                 else

 66                 {

 67                     next.step = cur.step + 2;

 68                 }

 69                 next.key = cur.key;

 70                 next.s = cur.s | bin[tmp-'A'];

 71             }

 72             else

 73             {

 74                 next.key = cur.key;

 75                 next.step = cur.step+1;

 76                 next.s = cur.s;

 77             }

 78             int dptmp = dp[next.x][next.y][next.key][next.s];

 79             if( dptmp < 0 )

 80             {

 81                 dp[next.x][next.y][next.key][next.s] = next.step;

 82                 qu.push(next);

 83             }

 84             else if(dptmp > next.step)

 85             {

 86                 dp[next.x][next.y][next.key][next.s] = next.step;

 87                 qu.push(next);

 88             }

 89         }

 90     }

 91     if( res == INT_MAX )

 92     {

 93         printf("impossible\n");

 94         return;

 95     }

 96     printf("%d\n", res);

 97 }

 98 int main(int argc, char *argv[])

 99 {

100     char tmp[MAXN+1];

101     while(1)

102     {

103         scanf("%d%d", &n, &m);

104         if( n == 0 && m == 0 ) return 0;

105         int sn = 0;

106         for( int i = 0 ; i < n ; i++ )

107         {

108             scanf("%s", tmp); 

109             for( int j = 0 ; j < n ; j++ )

110             {

111                 if( tmp[j] == 'K' )

112                 {

113                     start.x = i;

114                     start.y = j;

115                     start.key = 0;

116                     start.step = 0;

117                     start.s = 0;

118                 }

119                 else if( tmp[j] == 'T' )

120                 {

121                     ed.x = i;    

122                     ed.y = j;

123                 }

124                 else if( tmp[j] == 'S' )

125                 {

126                     map[i][j] = 'A'+sn;

127                     sn++;

128                     continue;

129                 }

130                 map[i][j] = tmp[j];

131             }

132         }

133         solve();

134     }

135 }

View Code