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hdu5025Saving Tang Monk（bfs+优先队列+状态压缩）

2024-07-23 20:12:36   218人阅读

题目链接：

huangjing

题意：

给了一幅图，然后这幅图里面有一个孙悟空，一个唐神，然后还有m把钥匙，还有最多5条蛇，然后蛇只要第一次杀死，杀这条蛇的时间为1S，并且后来再遇到的时候就不用加时间了，求最小的拯救时间。
思路：
首先那5条蛇可以用5位二进制数表示，然后开一个3维的数组判重，前两维保存坐标，后一维保存钥匙的状态，那么就可以了，还要注意的是因为有杀蛇这个操作，所以用优先队列来维护每一次搜索的都是最小的步数。。那么这个问题就解决了，我开始wa到死，就是在判断是否是蛇的时候，我判断如果当前可以杀蛇的话，那么就可以时间+2，但是我没有判断这条蛇已经被杀的话，时间只需加1，所以一直wa。。对着别人代码敲最后才发现。。。

题目：

Saving Tang Monk

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 471    Accepted Submission(s): 178


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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代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int dx[]={-1,1,0,0};
int dy[]={0,0,-1,1};

const int maxn=100+10;
int n,m;
char mp[maxn][maxn];

struct Node
{
    int x,y;
    int key,time,snake;
    friend bool operator<(Node n1,Node n2)
    {
        return n2.time<n1.time;
    }
};

priority_queue<Node>Q;

Node st,en;

bool vis[maxn][maxn][11];
int snake[maxn][maxn];

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

int judge(Node p,int dd)
{
    if(p.snake&(1<<dd))
        return 0;
    return 1;
}

int bfs()
{
   int xx,yy;
   while(!Q.empty())  Q.pop();
   Node cur,next;
   memset(vis,false,sizeof(vis));
   vis[st.x][st.y][0]=true;
   Q.push(st);
   while(!Q.empty())
   {
       cur=Q.top();
       Q.pop();
       if(cur.x==en.x&&cur.y==en.y&&cur.key==m)
            return cur.time;
       for(int i=0;i<4;i++)
       {
           xx=cur.x+dx[i];
           yy=cur.y+dy[i];
           if(check(xx,yy)&&!vis[xx][yy][cur.key])
           {
                if(mp[xx][yy]==cur.key+1+'0')
                {
                    vis[xx][yy][cur.key+1]=true;//񈬀
                    next.x=xx,next.y=yy;
                    next.snake=cur.snake;
                    next.key=cur.key+1;
                    next.time=cur.time+1;
                    Q.push(next);
                }
                else if(mp[xx][yy]=='S')//Óöµ½Éß
                {
                   if(!(cur.snake&(1<<snake[xx][yy])))
                   {
                     next.x=xx,next.y=yy;
                     next.key=cur.key;
                     next.snake=cur.snake|(1<<snake[xx][yy]);
                     next.time=cur.time+2;
                     vis[xx][yy][next.key]=true;
                     Q.push(next);
                   }
                   else
                    {
                        next.x=xx,next.y=yy;
                        next.key=cur.key;
                        next.snake=cur.snake;
                        next.time=cur.time+1;
                        vis[xx][yy][next.key]=true;
                        Q.push(next);
                    }
                }
               else
               {
                   next.x=xx,next.y=yy;
                   next.key=cur.key;
                   next.snake=cur.snake;
                   next.time=cur.time+1;
                   vis[xx][yy][next.key]=true;
                   Q.push(next);
               }
           }
       }
   }
   return -1;
}

void Read_Graph()
{
    char str[maxn];
    memset(snake,-1,sizeof(snake));
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%s",str+1);
        for(int j=1;j<=n;j++)
        {
            mp[i][j]=str[j];
            if(mp[i][j]=='K')
            {
                st.x=i;
                st.y=j;
                st.key=0;
                st.time=0;
                st.snake=0;
            }
            else if(mp[i][j]=='T')
            {
                en.x=i;
                en.y=j;
                en.key=m;
            }
            else if(mp[i][j]=='S')
              snake[i][j]=cnt++;
        }
    }
}

int main()
{

     while(~scanf("%d%d",&n,&m))
     {
        if(n==0&&m==0)  return 0;
        Read_Graph();
        int ans=bfs();
        if(ans==-1)
            printf("impossible\n");
        else
            printf("%d\n",ans);
     }
    return 0;
}

/*
3
3 2
K#T
.S.
21.
*/


 

hdu5025Saving Tang Monk（bfs+优先队列+状态压缩）


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