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POJ 5025 Saving Tang Monk（狀壓搜索）

編輯：C++入門知識

POJ 5025 Saving Tang Monk（狀壓搜索）

搜索題：注意蛇的狀態（第一次路過要殺掉蛇花2s，第二次以後1s）。狀態為坐標，蛇加鑰匙最大值。用【坐標加拿到最高等級的鑰匙】去重,不用蛇狀態去重因為不可能專門去殺一條蛇在同一地方呆兩次卻有可能專門去拿鑰匙。

代碼：

#include
#include
#include
#include
#include
#include
#include
#include
#define MOD 1000000007
typedef long long ll;
using namespace std;

const int maxn=105;
char g[maxn][maxn];
bool done[maxn][maxn][10];
int sn[maxn][maxn];
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
int n,m,sx,sy,ex,ey;
struct node
{
    int x,y,w,m,s;
    node(int xx=0,int yy=0,int ww=0,int mm=0,int ss=0)//m為拿到的最大鑰匙編號，s為蛇的壓縮狀態1~2^5
    {
        x=xx;y=yy;w=ww;m=mm;s=ss;
    }
    bool operator<(const node &a)const
    {
        return w>a.w;
    }
};
bool is_ill(int x,int y)
{
    return x<0||y<0||x>=n||y>=n;
}
int bfs()
{
    priority_queueq;
    memset(done,0,sizeof done);
    q.push(node(sx,sy,0,0,0));
    done[sx][sy][0]=1;
    while(!q.empty())
    {
        node e=q.top();q.pop();
        //printf(%d %d %d %d %d
,e.x,e.y,e.w,e.m,e.s);
        for(int i=0;i<4;i++)
        {
            int curx=e.x+dx[i];
            int cury=e.y+dy[i];
            if(is_ill(curx,cury)||g[curx][cury]=='#')continue;
            if(curx==ex&&cury==ey&&e.m>=m)return e.w+1;
            if(g[curx][cury]=='.'&&!done[curx][cury][e.m])
            {
                q.push(node(curx,cury,e.w+1,e.m,e.s));
                done[curx][cury][e.m]=1;
            }
            if(g[curx][cury]=='S'&&!done[curx][cury][e.m])
            {
                if((e.s&sn[curx][cury])==0)//沒殺過該位置的蛇
                {
                    q.push(node(curx,cury,e.w+2,e.m,e.s|sn[curx][cury]));
                    done[curx][cury][e.m]=1;
                }
                else
                {
                    q.push(node(curx,cury,e.w+1,e.m,e.s));
                    done[curx][cury][e.m]=1;
                }
            }
            if(g[curx][cury]>='0'&&g[curx][cury]<='9')
            {
                if(e.m+1==g[curx][cury]-'0'&&!done[curx][cury][e.m+1])//拿到下一個鑰匙
                {
                    q.push(node(curx,cury,e.w+1,e.m+1,e.s));
                    done[curx][cury][e.m+1]=1;
                }
                else if(!done[curx][cury][e.m])//最好拿到下一個嘛，拿不到直接走過去也行。（鑰匙太高太低都不管，並且不可能有鑰匙不拿（指同時拓展拿的和不拿的情況））
                {
                    q.push(node(curx,cury,e.w+1,e.m,e.s));
                    done[curx][cury][e.m]=1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    while(~scanf(%d%d,&n,&m)&&(n||m))
    {
        int cnt=0;
        memset(sn,0,sizeof sn);
        for(int i=0;i 

Saving Tang Monk
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 149    Accepted Submission(s): 57






Problem Description

《Journey to the West》(also 《Monkey》) is one of the Four Great Classical Novels of Chinese literature. It was written by Wu Cheng'en during the Ming Dynasty. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha Wujing, escorted Tang Monk to India to
 get sacred Buddhism texts. 



During the journey, Tang Monk was often captured by demons. Most of demons wanted to eat Tang Monk to achieve immortality, but some female demons just wanted to marry him because he was handsome. So, fighting demons and saving Monk Tang is the major job for
 Sun Wukong to do.



Once, Tang Monk was captured by the demon White Bones. White Bones lived in a palace and she cuffed Tang Monk in a room. Sun Wukong managed to get into the palace. But to rescue Tang Monk, Sun Wukong might need to get some keys and kill some snakes in his way.



The palace can be described as a matrix of characters. Each character stands for a room. In the matrix, 'K' represents the original position of Sun Wukong, 'T' represents the location of Tang Monk and 'S' stands for a room with a snake in it. Please note that
 there are only one 'K' and one 'T', and at most five snakes in the palace. And, '.' means a clear room as well '#' means a deadly room which Sun Wukong couldn't get in.



There may be some keys of different kinds scattered in the rooms, but there is at most one key in one room. There are at most 9 kinds of keys. A room with a key in it is represented by a digit(from '1' to '9'). For example, '1' means a room with a first kind
 key, '2' means a room with a second kind key, '3' means a room with a third kind key... etc. To save Tang Monk, Sun Wukong must get ALL kinds of keys(in other words, at least one key for each kind).



For each step, Sun Wukong could move to the adjacent rooms(except deadly rooms) in 4 directions(north, west, south and east), and each step took him one minute. If he entered a room in which a living snake stayed, he must kill the snake. Killing a snake also
 took one minute. If Sun Wukong entered a room where there is a key of kind N, Sun would get that key if and only if he had already got keys of kind 1,kind 2 ... and kind N-1. In other words, Sun Wukong must get a key of kind N before he could get a key of
 kind N+1 (N>=1). If Sun Wukong got all keys he needed and entered the room in which Tang Monk was cuffed, the rescue mission is completed. If Sun Wukong didn't get enough keys, he still could pass through Tang Monk's room. Since Sun Wukong was a impatient
 monkey, he wanted to save Tang Monk as quickly as possible. Please figure out the minimum time Sun Wukong needed to rescue Tang Monk.

 



Input

There are several test cases.



For each case, the first line includes two integers N and M(0 < N <= 100, 0<=M<=9), meaning that the palace is a N×N matrix and Sun Wukong needed M kinds of keys(kind 1, kind 2, ... kind M). 



Then the N × N matrix follows.



The input ends with N = 0 and M = 0.

 



Output

For each test case, print the minimum time (in minutes) Sun Wukong needed to save Tang Monk. If it's impossible for Sun Wukong to complete the mission, print impossible(no quotes).

 



Sample Input3 1
K.S
##1
1#T
3 1
K#T
.S#
1#.
3 2
K#T
.S.
21.
0 0


Sample Output
5
impossible
8


Source 2014 ACM/ICPC Asia Regional Guangzhou Online

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