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[4][2]={{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
char map[MAXN][MAXN];
int bin[] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096};
int dp[MAXN][MAXN][10][35];
int n, m;
class state
{
    public:
        int x, y, step, key, s;
};
state start, ed;
void solve()
{
    memset(dp, -1, sizeof(dp));
    queue<state> qu; 
    qu.push(start);
    int res = INT_MAX;
    while( qu.size() > 0 )
    {
        state cur = qu.front();
        qu.pop();
        for( int i = 0 ; i < 4 ; 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][0];
            next.y = cur.y+dir[i][1];
            next.key = cur.key;
            next.s = cur.s;
            next.step = cur.step;
            if( next.x < 0 || next.x >= n || next.y < 0 || next.y >= n )//出界
            {
                continue;
            }
            char tmp = map[next.x][next.y];
            if( tmp  == '#' )//陷阱
            {
                continue;
            }
            if( tmp >= '1' && tmp <= '9' && cur.key >= tmp-'0'-1 )//有钥匙
            {
                next.step = cur.step+1;
                next.key = max(cur.key, tmp - '0');
                next.s = cur.s;
            }
            else if( tmp >= 'A' && tmp <= 'F')//蛇
            {
                if( cur.s/bin[tmp-'A']%2 == 1 )
                {
                    next.step = cur.step + 1;    
                }
                else
                {
                    next.step = cur.step + 2;
                }
                next.key = cur.key;
                next.s = cur.s | bin[tmp-'A'];
            }
            else
            {
                next.key = cur.key;
                next.step = cur.step+1;
                next.s = cur.s;
            }
            int dptmp = dp[next.x][next.y][next.key][next.s];
            if( dptmp < 0 )
            {
                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
");
        return;
    }
    printf("%d
", res);
}
int main(int argc, char *argv[])
{
    char tmp[MAXN+1];
    while(1)
    {
        scanf("%d%d", &n, &m);
        if( n == 0 && m == 0 ) return 0;
        int sn = 0;
        for( int i = 0 ; i < n ; i++ )
        {
            scanf("%s", tmp); 
            for( int j = 0 ; j < n ; j++ )
            {
                if( tmp[j] == 'K' )
                {
                    start.x = i;
                    start.y = j;
                    start.key = 0;
                    start.step = 0;
                    start.s = 0;
                }
                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();
    }
}