POJ 5025 Saving Tang Monk（状压搜索）

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

搜索题：注意蛇的状态（第一次路过要杀掉蛇花2s，第二次以后1s）。状态为坐标 ，蛇 加钥匙最大值。用【坐标加拿到最高等级的钥匙】去重,不用蛇状态去重因为不可能专门去杀一条蛇在同一地方呆两次却有可能专门去拿钥匙。

代码：

#include<iostream>
#include<queue>
#include<map>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#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为拿到的最大钥匙编号，s为蛇的压缩状态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_queue<node>q;
    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\n",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)//没杀过该位置的蛇
                {
                    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])//拿到下一个钥匙
                {
                    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])//最好拿到下一个嘛，拿不到直接走过去也行。（钥匙太高太低都不管，并且不可能有钥匙不拿（指同时拓展拿的和不拿的情况））
                {
                    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<n;i++)
        {
            scanf("%s",g[i]);
            for(int j=0;j<n;j++)
                if(g[i][j]=='K')sx=i,sy=j,g[i][j]='.';
                else if(g[i][j]=='T')ex=i,ey=j,g[i][j]='.';
                else if(g[i][j]=='S')sn[i][j]=1<<(cnt++);
        }
        int ans=bfs();
        if(ans)printf("%d\n",ans);
        else printf("impossible\n");
    }
    return 0;
}

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 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

 

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