[ACM] HDU 5025 Saving Tang Monk （状态压缩，BFS）

Saving Tang Monk

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 941    Accepted Submission(s): 352

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 Input

3 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

解题思路：

题意为一个地图，'K'代表孙悟空的位置，也就是起点，'T'代表唐僧的位置，数字‘1’‘2’等代表第几种钥匙，'S'代表蛇，'#'不能走，题意的目的就是孙悟空去救唐僧，要求前提必须是拿到给定的m种钥匙，才干去救唐僧，除了'#‘的位置，其它位置都能够走（假设到达了唐僧的位置，但没拿到给定的m种钥匙，任务也没法完毕，必须得拿到m钟钥匙)，到达'S'位置，要多花一分钟杀死蛇，其它位置走一步花一分钟，问最少花多少分钟才干挽救唐僧，假设不能，输出impossible.

 题意真的不好理解：要想挽救唐僧，仅仅有唯一的方法，就是取得全部种类的钥匙，然后到达唐僧的位置，去挽救。孙悟空位置和唐僧位置能够走多次。

还有蛇的状态也须要特别注意，第一次杀死蛇，第二次再到达该位置时，就不用再杀蛇了，给蛇编号，杀了为1，不杀为0，状态压缩。

定义 f[i][j][k][state] ,为坐标位置走到坐标 i, j ，时已经取得了第k种钥匙，当前蛇的状态为state。 进行广度优先搜索。

參考：http://www.cnblogs.com/whatbeg/p/3983522.html

代码：

#include <iostream>
#include <algorithm>
#include <string.h>
#include <map>
#include <queue>
using namespace std;

const int inf=0x3f3f3f3f;
int n,m;
char mp[102][102];
int dx[4]={0,0,-1,1};
int dy[4]={1,-1,0,0};
int f[102][102][12][35];//f[i][j][k][state]为坐标i,j位置处已经拿到第k种钥匙且杀死蛇的状态为state时的最小步数
int sx,sy;//開始位置
int scnt;//蛇的数量
int ans;
map<pair<int,int>,int>snake;//给某一位置上的蛇编号,从1開始

struct Node
{
    int x,y,k,s,step;
};
queue<Node>q;


void init()
{
    memset(f,inf,sizeof(f));
    snake.clear();
    scnt=0;
    ans=inf;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        cin>>mp[i][j];
        if(mp[i][j]=='K')
            sx=i,sy=j,mp[i][j]='.';//别忘了mp[i][j]='.'这一句，起始位置也能够反复多次走
        else if(mp[i][j]=='S')
            snake[make_pair(i,j)]=++scnt;//编号
    }
    f[sx][sy][0][0]=0;
}


bool ok(int x,int y)
{
    if(x>=1&&x<=n&&y>=1&&y<=n&&mp[x][y]!='#')
        return true;
    return false;
}


void bfs(Node st)
{
    if(!q.empty())
        q.pop();
    q.push(st);
    while(!q.empty())
    {
        Node cur=q.front();
        q.pop();
        Node next;//下一步的状态节点
        for(int i=0;i<4;i++)
        {
            int newx=cur.x+dx[i];
            int newy=cur.y+dy[i];
            if(!ok(newx,newy))
                continue;
            next.x=newx,next.y=newy;
            if(mp[newx][newy]=='S')
            {
                int th=snake[make_pair(newx,newy)];
                if(cur.s&(1<<(th-1)))//已经杀过该条蛇
                {
                    next.s=cur.s;
                    next.k=cur.k;
                    next.step=cur.step+1;
                }
                else
                {
                    next.s=(cur.s|(1<<(th-1)));
                    next.k=cur.k;
                    next.step=cur.step+2;
                }
                if(next.step<f[newx][newy][next.k][next.s])
                {
                    f[newx][newy][next.k][next.s]=next.step;
                    q.push(next);
                }
            }
            else if(mp[newx][newy]>='1'&&mp[newx][newy]<='9')
            {
                int th=mp[newx][newy]-'0';
                if(th==cur.k+1)//当前这个钥匙正好是想要的，已有钥匙加1
                    next.k=cur.k+1;
                else
                    next.k=cur.k;
                next.s=cur.s;
                next.step=cur.step+1;
                if(next.step<f[newx][newy][next.k][next.s])
                {
                    f[newx][newy][next.k][next.s]=next.step;
                    q.push(next);
                }
            }
            else if(mp[newx][newy]=='.')
            {
                next.k=cur.k;
                next.s=cur.s;
                next.step=cur.step+1;
                if(next.step<f[newx][newy][next.k][next.s])
                {
                    f[newx][newy][next.k][next.s]=next.step;
                    q.push(next);
                }
            }
            else if(mp[newx][newy]=='T')
            {
                next.k=cur.k;
                next.s=cur.s;
                next.step=cur.step+1;
                if(next.step<f[newx][newy][next.k][next.s])
                    f[newx][newy][next.k][next.s]=next.step;
                if(next.k==m)//该状态下不再向下扩展
                    ans=min(f[newx][newy][next.k][next.s],ans);
                else
                    q.push(next);
            }
        }
    }
}

int main()
{
    while(cin>>n>>m&&(n||m))
    {
        init();
        Node temp;
        temp.x=sx,temp.y=sy,temp.k=0,temp.s=0,temp.step=0;
        bfs(temp);
        if(ans==inf)
            cout<<"impossible"<<endl;
        else
            cout<<ans<<endl;
    }
    return 0;
}