HDU5025(状压搜索)

Saving Tang Monk

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2550 Accepted Submission(s): 897

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

题意：给你一个矩阵，里面有一个起点，一个终点，然后还有若干种钥匙，和蛇，要得到第N种钥匙必须先得到1~N- 1种钥匙每一种的至少一把，然后如果遇到蛇会多用一个时间来消灭蛇，问你最少需要多少时间到达终点。

解题思路：状压bfs，开思维就行，分别表示横坐标，纵坐标，钥匙状态，蛇的状态。注意蛇的状态特殊处理一下就行。具体思路看代码。

#include<bits/stdc++.h>using namespace std;const int maxn = 102;int N, M;int sx, sy;//起点int tx, ty;//终点char Matrix[maxn][maxn];//地图bool visit[maxn][maxn][12][32];int cnt;int dx[4] = {0, 1, 0, -1};int dy[4] = {1, 0, -1, 0};int snake;struct sna{    int x, y;}Sna[10];struct node{    int x, y, step, status, s;    node(){        step = s = 0;        status = 0;    }};bool ok(int status,int id){    if(status + 1 == id) return true;    else return false;}int bfs(){    queue<node> Q;    while(!Q.empty()) Q.pop();    node n0, nx, ans;    n0.x = sx;    n0.y = sy;    Q.push(n0);    while(!Q.empty())    {        nx = Q.front();        Q.pop();        int x = nx.x;        int y = nx.y;        int step = nx.step;        int status = nx.status;        int s = nx.s;        if(x == tx && y == ty && ok(status, M + 1)) return step;        visit[x][y][status][s] = true;        bool flag = false;        if(Matrix[x][y] == 'S')        {            int id;            for(int i = 0; i < cnt; i++)            {                if(x == Sna[i].x && y == Sna[i].y)                {                    id = i;                    break;                }            }            int xx = (s>>id)&1;            if(xx == 1)            {                flag = true;            }            else            {                int v = (1<<id);                int _s = s|v;                ans.x = x;                ans.y = y;                ans.status = status;                ans.s = _s;                ans.step = step + 1;                Q.push(ans);            }        }        else flag = true;        for(int i = 0; i < 4 && flag; i++)        {            int _x = x + dx[i];            int _y = y + dy[i];            if(_x >= 1 && _x <= N && _y >= 1 && _y <= N)            {                if(Matrix[_x][_y] >= '1' && Matrix[_x][_y] <= '9')//钥匙                {                    int id = Matrix[_x][_y] - '0';                    if(ok(status,id))                    {                        int _status = status + 1;                        if(!visit[_x][_y][_status][s])                        {                            visit[_x][_y][_status][s] = true;                            ans.x = _x;                            ans.y = _y;                            ans.step = step + 1;                            ans.status = _status;                            ans.s = s;                            Q.push(ans);                        }                    }                    else                    {                        if(!visit[_x][_y][status][s])                        {                            visit[_x][_y][status][s] = true;                            ans.x = _x;                            ans.y = _y;                            ans.s = s;                            ans.step = step + 1;                            ans.status = status;                            Q.push(ans);                        }                    }                }                else if(Matrix[_x][_y] != '#')                {                    if(!visit[_x][_y][status][s])                    {                        ans.x = _x;                        ans.y = _y;                        ans.status = status;                        ans.step = step + 1;                        ans.s = s;                        visit[_x][_y][status][s] = true;                        Q.push(ans);                    }                }            }        }    }    return -1;}int main(){    while(~scanf("%d%d", &N, &M))    {        if(N == 0 && M == 0) break;        memset(visit,false,sizeof(visit));        cnt = 0;        for(int i = 1; i <= N; i++)        {            scanf("%s", Matrix[i] + 1);            for(int j = 1; j <= N; j++)            {                if(Matrix[i][j] == 'K')                {                    sx = i;                    sy = j;                }                if(Matrix[i][j] == 'T')                {                    tx = i;                    ty = j;                }                if(Matrix[i][j] == 'S')                {                    Sna[cnt].x = i;                    Sna[cnt++].y = j;                }            }        }        int res = bfs();        if(res != -1) printf("%d\n",res);        else printf("impossible\n");    }    return 0;}