Hdu 6171 --- Admiral (bidirectional search), hdu6171 --- admiral

Hdu 6171 --- Admiral (bidirectional search), hdu6171 --- admiral

Question Link

 

Problem DescriptionSuppose that you are an admiral of a famous naval troop. Our naval forces have got 21 battleships. There are 6 types of battleships.
First, we have got one flagship in which the admiral must be and it is denoted by number 0. others are denoted by number from 1 to 5, each of them has 2, 3, 4, 5, 6 ships of its kind. so, we have got 21 battleships in total and we must take a giant battle against the enemy. hence, the correct strategy of how to arrange each type of battleships is very important to us.
The shape of the battlefield is like the picture that is shown below.
To simplify the problem, we consider all battleships have the same rectangular shape.

Fortunately, we have already known the optimal state of battleships.
As you can see, the battlefield consists of 6 rows. and we have 6 types of battleship, so the optimal state is that all the battleships denoted by number I are located at the I-th row. hence, each type of battleship corresponds to different color.
You are given the initial state of battlefield as input. You can change the state of battlefield by changing the position of flagship with adjacent battleship.
Two battleships are considered adjacent if and only if they are not in the same row and share parts of their edges. for example, if we denote the cell which is at I-th row and j-th position from the left as (I, j), then the cell (2, 1) is adjacent to the cells ).
Your task is to change the position of the battleships minimum times so as to reach the optimal state.
Note: All the coordinates are 0-base indexed.

 

InputThe first line of input contains an integer T (1 <= T <= 10), the number of test cases.
Each test case consists of 6 lines. The I-th line of each test case contains I integers, denoting the type of battleships at I-th row of battlefield, from left to right.

 

OutputFor each test case, if you can't reach the goal in no more than 20 moves, you must output "too difficult" in one line. otherwise, you must output the answer in one line.

 

Sample Input112 02 1 23 3 3 34 4 4 45 5 5 5

 

Sample Output3. There are 21 numbers composed of 1 0, 2 1, 3 2, 4 3, 5 4, and 6 5. The 21 numbers constitute a triangle, just like the Yang Hui triangle, the number of the first row is 1, and the number of the second row is 2 ...... There are 6 numbers in the 6th rows. Now we need to set the 21 numbers in order. 0 is in the first row, and 1 is in the second row ...... 5 In the sixth row, each operation is 0 and the adjacent number can be exchanged, and the exchange cannot be performed between the same row. Find the minimum number of steps to normalize all the numbers, if I still cannot complete the 20 steps, I will output "too difficult ". Idea: search, but direct search for 20 steps will time out, so you can search for 10 steps from both ends, which reduces the complexity, and you can use hash to save the status. The Code is as follows:

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <map>using namespace std;typedef long long LL;int dx[4]={1,1,-1,-1};int dy[4]={0,1,-1,0};struct Node{    LL p[6][6];    int r,c;    int flag;    int dept;};queue<Node>Q;map<LL,int>M[2];LL cal(Node a){    LL ans=0;    for(int i=0;i<6;i++)    {        for(int j=0;j<=i;j++)        {            ans=ans*6+a.p[i][j];        }    }    return ans;}int bfs(Node &s,Node &e){    while(!Q.empty()) Q.pop();    M[0].clear(); M[1].clear();    M[0][cal(s)]=0;    M[1][cal(e)]=0;    Q.push(s);    Q.push(e);    while(!Q.empty())    {        Node x=Q.front(); Q.pop();        LL sta=cal(x);        if(M[!x.flag].count(sta))        {            int num=M[!x.flag][sta]+x.dept;            if(num<=20) return num;            else continue;        }        if(x.dept>=10) continue;        for(int i=0;i<4;i++)        {            Node y=x;            y.dept++;            y.r+=dx[i];            y.c+=dy[i];            if(y.r<0 || y.r>=6 || y.c<0 || y.c>y.r) continue;            swap(y.p[x.r][x.c],y.p[y.r][y.c]);            if(M[y.flag].count(cal(y))==0) M[y.flag][cal(y)]=y.dept;            Q.push(y);        }    }    return -1;}int main(){    int T; cin>>T;    Node s,e;    while(T--)    {        for(int i=0;i<6;i++)        {            for(int j=0;j<=i;j++)            {                scanf("%lld",&s.p[i][j]);                if(s.p[i][j]==0) s.r=i, s.c=j;                e.p[i][j]=i;            }        }        s.flag=0; s.dept=0;        e.r=0; e.c=0;        e.flag=1; e.dept=0;        int ans=bfs(s,e);        if(ans>=0&&ans<=20) printf("%d\n",ans);        else puts("too difficult");    }    return 0;}/**12 12 0 23 3 3 34 4 4 4 45 5 5 5 5 501 12 2 23 3 3 34 4 4 4 45 5 5 5 5 5*/