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:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where 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:
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.
输入
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
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.
样例输入
2 3 4 1 5 x 7 6 8
样例输出
ullddrurdllurdruldr
来源
South Central USA 1998
#include <iostream>
#include <vector>
#include <queue>
#include <set>
#include <cstring>
#include <cstdlib>
using namespace std;//一共9!状态,012345678
struct status {
char a[9];
vector<char> path;
};
queue<status> q;
set<int> st;//vis[]
void newStatus(status last, int otherpos, int zeropos, char c)
{
if (otherpos >= 0 && otherpos <= 8) {
swap(last.a[otherpos], last.a[zeropos]);
status newS;
if (st.find(atoi(last.a)) == st.end()) {
newS = last;
newS.path.push_back(c);//路径
q.push(newS);//入队
st.insert(atoi(last.a));//vis
}
}
}
int main()
{
status initialStatus;
char Input[50];
initialStatus.path.clear();
while (cin.getline(Input, 48)) {
bool flag = false;
int j = 0;
for (int i = 0; Input[i]; i++)
if (Input[i] != ' ')
if (Input[i] == 'x')
initialStatus.a[j++] = '0';
else
initialStatus.a[j++] = Input[i];
st.clear();
while (q.size() != 0)
q.pop();
q.push(initialStatus);
st.insert(atoi(initialStatus.a));
while (q.size() != 0) {
status tmp;
tmp = q.front();
if (atoi(tmp.a) == 123456780) {
for (int i = 0; i < tmp.path.size(); i++)
cout << tmp.path[i];
cout << endl;
flag = true;
break;
}
q.pop();
int zeropos;
for (int i = 0; i < 9; i++)
if (tmp.a[i] == '0')
zeropos = i;
newStatus(tmp, zeropos - 3, zeropos, 'u');
newStatus(tmp, zeropos + 3, zeropos, 'd');
if (zeropos % 3 != 0)
newStatus(tmp, zeropos - 1, zeropos, 'l');
if (zeropos % 3 != 2)
newStatus(tmp, zeropos + 1, zeropos, 'r');
}
if (flag == false)
cout << "unsolvable" << endl;
}
}
过不了,为什么,感觉没问题,试了几个测试用例跟网上通过了的答案都一样,路径这样记录有啥问题吗。