1015 八数码 bfs
题目描述
The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:
这个第十五道题，已经存在了一百年以上了，即使你不知道它的名字，你一定也见过它。它由十五个浮动瓷砖组成，每个瓷砖有一个1到15的数字，并且这些瓷砖被塞进一个4*4的方块里，方块少了一个瓷砖。让我们叫这个少的瓷砖为‘x’，这个问题是让这些瓷砖移动使它有序。比如：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 xwhere the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
仅有一种合法操作使'x' 与它相连的唯一的瓷砖交换。例如，下面的操作解决了一个略显复杂的谜题。
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
前一行中的字母表示在每一步中'x' 瓷砖的哪个邻居与'x' 交换;合法的值是'r','l','u' and 'd',对应右左上和下。
并且并非所有谜题都能解决；1870年，一个叫做山姆罗伊德因发行了一个无法解决的谜题版本而闻名，并打击了很多人。事实上，要让一个普通谜题变成无法解决的谜题，只需要交换两个贴图。
（不包括x贴图）。
输入描述:
You will receive a description of a configuration of the 8 puzzle. The description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
输出描述:
You will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line.
示例1
输出
复制ullddrurdllurdruldr
分析
用map存每个状态对应的改变
然后bfs每一个状态
可以交换字符串中的两个值来改变当前的字符串
通过除以长度，和模长度得到当前位置在9*9方格中的相对坐标
//-------------------------代码----------------------------
#define int LL
const int N = 1e5+10;
int n,m;
string s;
int a[100];
unordered_map<string,string> mp;
void bfs() {
queue<string>q;
q.push(s); //初始状态
mp[s] = "";
string ed = "12345678x"; //结束状态
while(q.size()) {
auto t = q.front();q.pop();
if(t == ed) {cout<<mp[t];rt;}
int pos = t.find('x'); //找到当前字符串中x的位置
int x = pos / 3, y = pos % 3; //一维位置转二维坐标
string dt = mp[t];
for(int i = 0;i<4;i++) {
int xx = x + dx[i],yy = y + dy[i];
if(xx < 0 || yy < 0 || xx >= 3 || yy >= 3) continue;
string s1 = t;
swap(s1[xx * 3 + yy],s1[pos]);
if(mp.count(s1)) continue; //不小心把count中间的值写成了t
mp[s1] = dt + dc[i];
q.push(s1);
}
}
cout<<"unsolvable";
}
void solve()
{
s = "";
for(int i = 1;i<=9;i++) {//*忘了输入要一个个输入了
char c;
cin>>c;
s += c;
}
bfs();
}
signed main(){
TLE;
// int t;cin>>t;while(t -- )
solve();
// {solve(); }
return 0;
}
/*样例区
*/
//------------------------------------------------------------
忘了字符串是一个字符一个字符输入的直接cin>>s 了。。
不小心把map 的count 里的变量写成了t 这个变换前的变量。。
