2017多校联合第十场／HDU 6171 Admiral 双向bfs + 哈希
Admiral
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 153428/153428 K (Java/Others)Total Submission(s): 334 Accepted Submission(s): 105
Problem Description
Suppose 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 (1,0), (1,1), (3,1), (3,2).
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.
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 (1,0), (1,1), (3,1), (3,2).
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.
Input
The 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.
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.
Output
For 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 Input
1 1 2 0 2 1 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 5
Sample Output
3
Source
2017 Multi-University Training Contest - Team 10
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题意：给你一个高度为6的数塔，问最少多少步能走成如下状态。
每次只能将值为0的点与其左上方，上方，下方，右下方四个方向的一个点进行交换。
0
1 1
2 2 2
3 3 3 3
4 4 4 4 4
5 5 5 5 5 5
大于20步可以直接输出“too difficult”
题解：一共21个点，每个点有6个状态，一共6^21个状态，所以状态压缩数组是开不下的，所以用hash，long long可以存的下。已知初态和终态，所以要双向bfs。先预处理终态，预处理10步，然后看初态十步只能能否和预处理的结果一样，可以用map维护，一样则输出step1+step2，否则输出too difficult。
代码：
typedef unsigned long long ull;
struct node
{
int val[6][6];
int r,c;//0的位置
bool tp;
int step;
};
ull Hash(node &t)
{
ull tmp=0;
for(int i=0;i<6;i++)
{
for(int j=0;j<=i;j++)
tmp=tmp*6+t.val[i][j];
}
return tmp;
}
mapbook[2];
queueq;
int dx[]={-1,-1,1,1};
int dy[]={-1,0,0,1};
int bfs(node &s,node &t)
{
while(!q.empty()) q.pop();
book[0].clear();
book[1].clear();
s.tp=0;
t.tp=1;
s.step=t.step=0;
book[s.tp][Hash(s)]=0;
book[t.tp][Hash(t)]=0;
q.push(s);
q.push(t);
while(!q.empty())
{
node x,a=q.front();
q.pop();
ull tmp=Hash(a);
if(book[!a.tp].count(tmp))
if(book[!a.tp][tmp]+a.step<=20)
return book[!a.tp][tmp]+a.step;
else continue;
if(a.step>=10) continue;
for(int i=0;i<4;i++)
{
x=a;
x.r+=dx[i];
x.c+=dy[i];
if(x.r>=6 || x.c>x.r || x.c<0 || x.r<0) continue;
swap(x.val[x.r][x.c],x.val[a.r][a.c]);
tmp=Hash(x);
if(book[x.tp].count(tmp)) continue;
book[x.tp][tmp]=++x.step;
q.push(x);
}
}
return -1;
}
int main()
{
ios::sync_with_stdio(false);
int T;
cin>>T;
node s,t;
while(T--)
{
for(int i=0;i<6;i++)
for(int j=0;j<=i;j++)
{
cin>>s.val[i][j];
if(s.val[i][j]==0) s.r=i,s.c=j;
t.val[i][j]=i;
}
t.r=t.c=0;
int ans=bfs(s,t);
ans==-1?cout<<"too difficult"<