hdoj 5025 Saving Tang Monk 【BFS+状态压缩】

Saving Tang Monk

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

Total Submission(s): 1851    Accepted Submission(s): 653

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 1K.S##11#T3 1K#T.S#1#.3 2K#T.S.21.0 0

 

Sample Output

5impossible8

 

题意：给定一个N*N的地图，上面有六种字符——'#'表示此处不能走。'.'表示此处可以走。 'K'表示起点。'T'表示终点。'S'表示此处有一条蛇。数字(1-9)表示此处有一把钥匙。要想获取钥匙2，必须先获取钥匙1。打死一条蛇花费时间1，每移动一步花费时间1，从位置(x, y)->可以上下左右移动。现在问能否到达终点，若可以输出最少时间花费，反之输出-1。

思路：用vis[x][y][digit][state]表示在位置(x, y)处拥有前digit把和打死蛇的状态。因为蛇最多为5条，可以考虑用状态压缩去记录。

注意：需要用优先队列保证每一步解的最优性。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (100+10)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;bool vis[MAXN][MAXN][20][(1<<6)];char str[MAXN][MAXN];struct Node{    int x, y, step, digit, state;    friend bool operator < (Node a, Node b){        return a.step > b.step;    }};int n, m;bool judge(int x, int y){    return x >= 0 && x < n && y >= 0 && y < n && str[x][y] != '#';}int BFS(int x, int y){    int Move[4][2] = {0,1, 0,-1, 1,0, -1,0};    priority_queue<Node> Q;    Node now, next;    now.x = x; now.y = y; now.digit = now.step = now.state = 0;    Q.push(now); CLR(vis, false);    vis[x][y][0][0] = true;    while(!Q.empty())    {        now = Q.top();        Q.pop();        if(str[now.x][now.y] == 'T' && now.digit == m)            return now.step;        for(int k = 0; k < 4; k++)        {            next.x = now.x + Move[k][0];            next.y = now.y + Move[k][1];            if(!judge(next.x, next.y))                continue;            if(str[next.x][next.y] >= 'a' && str[next.x][next.y] <= 'z')            {                next.state = now.state; next.digit = now.digit;                int num = str[next.x][next.y] - 'a';                if(next.state & (1<<num))                    next.step = now.step + 1;                else                    next.step = now.step + 2;                next.state |= (1<<num);            }            else if(str[next.x][next.y] >= '0' && str[next.x][next.y] <= '9')            {                next.state = now.state;                next.digit = now.digit;                next.step = now.step + 1;                int num = str[next.x][next.y] - '0';                if(now.digit + 1 == num)                    next.digit++;            }            else            {                next.state = now.state;                next.digit = now.digit;                next.step = now.step + 1;            }            if(!vis[next.x][next.y][next.digit][next.state])            {                vis[next.x][next.y][next.digit][next.state] = true;                Q.push(next);            }        }    }    return -1;}int main(){    while(scanf("%d%d", &n, &m), n||m)    {        int sx, sy;        int num = 0;        for(int i = 0; i < n; i++)        {            Rs(str[i]);            for(int j = 0; j < n; j++)            {                if(str[i][j] == 'K')                {                    sx = i;                    sy = j;                }                if(str[i][j] == 'S')                    str[i][j] = 'a' + num++;            }        }        int ans = BFS(sx, sy);        printf(ans != -1 ? "%d\n" : "impossible\n", ans);    }    return 0;}