HDU 5025 Saving Tang Monk 【状态压缩BFS】

任意门：http://acm.hdu.edu.cn/showproblem.php?pid=5025

Saving Tang Monk

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3242    Accepted Submission(s): 1127

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

题意概括：

给一个 N * N 的地图，孙悟空起点在 K ，唐僧起点在 T，数字 i 代表第 i 把钥匙，# 是毒气区不能进入， S 有蛇（需要杀死才能通过，只需要杀死一次）；

孙悟空要按顺序集齐 M 把钥匙才能解救师父，求最短的时间，如果没有输出“impossible”。

解题思路：

BFS找最短路，因为蛇最多只有5条，所以状态压缩判断哪条已杀，哪条未杀；如果未杀则当前需要多走一步，反则不用。

集钥匙只需要一个变量 cnt_key 记录已经集了多少把钥匙，那么接下俩可以拿的钥匙就是 第 cnt_key+1 把。

BFS需要一个 vis[ x ][ y ][ key ][ snake ] 来book一下状态，已经处理过的不再处理。

最后一个debug了很久的原因：多测试用例当 N M 都为 0 时（ ... &&（N+M））才结束，习惯性打了（ ... &&N && M），凉凉。

AC code:

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <queue>
 #include <cmath>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int MAXN = ;
 int nx[] = {-, , , };
 int ny[] = {, , -, };
 struct date
 {
     int x, y, t, key, snake;
     date(int _x = , int _y = , int _t = , int _key = , int _snake = ):x(_x), y(_y), t(_t), key(_key), snake(_snake){}
 };

 char mmp[MAXN][MAXN];
 bool vis[MAXN][MAXN][][];
 int N, M, cnt;
 int sx, sy, ex, ey;

 void solve()
 {
     memset(vis, false, sizeof(vis));
     int ans = INF;                                  //初始化步数
     queue<date> sq;
     sq.push(date(sx, sy, , , ));
     //vis[sx][sy] = true;
     date tp;
     while(!sq.empty()){
         tp = sq.front(), sq.pop();
         int x = tp.x, y = tp.y, key = tp.key, snake = tp.snake, t = tp.t;
         if(key == M && mmp[x][y] == 'T') ans = min(ans, t);                     //集齐钥匙并且到达终点
         if(vis[x][y][key][snake]) continue;    //状态已访问过
         vis[x][y][key][snake] = true;               //状态不重复
         for(int k = ; k < ; k++){
             int tx = x + nx[k], ty = y + ny[k];
             if(tx <  || ty <  || tx == N || ty == N || mmp[tx][ty] == '#') continue;

             date now = tp;

             if('A' <= mmp[tx][ty] && mmp[tx][ty] <= 'G'){
                 int s = mmp[tx][ty] - 'A';
                 if((<<s) & now.snake);               //蛇被被打了
                 else{                               //蛇没有被打
                     now.snake |= (<<s);
                     now.t++;
                 }
             }
             else if(mmp[tx][ty] - '' == now.key + ){    //捡钥匙，遇到当前可以捡的钥匙
                 now.key++;
             }
             now.t++;
             sq.push(date(tx, ty, now.t, now.key, now.snake));
         }
     }
     if(ans >= INF) puts("impossible");
     else printf("%d\n", ans);
 }

 int main()
 {
     while(~scanf("%d%d", &N, &M) && (N+M)){
         cnt = ;
         for(int i = ; i < N; i++){
             scanf("%s", &mmp[i]);
             for(int j = ; j < N; j++){
                 if(mmp[i][j] == 'K') sx = i, sy = j;
                 if(mmp[i][j] == 'S') mmp[i][j] = cnt+'A', cnt++;
             }
         }
         solve();
     }
     return ;
 }