HDU 1043 Eight (BFS·八数码·康托展开)
题意 输出八数码问题从给定状态到12345678x的路径
用康托展开将排列相应为整数 即这个排列在全部排列中的字典序 然后就是基础的BFS了
- #include <bits/stdc++.h>
- using namespace std;
- const int N = 5e5, M = 9;
- int x[4] = { -1, 1, 0, 0};
- int y[4] = {0, 0, -1, 1};
- int fac[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};
- int puz[N][M], nex[N], dir[N], vis[N], q[N];
- int getCantor(int a[]) //康托展开 将排列转化为整数
- {
- int ret = 0;
- for(int i = 0; i < M; ++i)
- {
- for(int j = i + 1; j < M; ++j)
- if(a[j] < a[i]) ret += fac[M - i - 1];
- }
- return ret;
- }
- void bfs()
- {
- int t[M] = {1, 2, 3, 4, 5, 6, 7, 8, 0};
- int id = getCantor(t);
- dir[id] = -1;
- memcpy(puz[id], t, sizeof(t));
- memset(vis, 0, sizeof(vis));
- int r, c, k, nr, nc, nk, nid;
- int front = 0, rear = 0;
- q[rear++] = id;
- vis[id] = 1;
- while(front < rear)
- {
- int id = q[front++];
- memcpy(t, puz[id], sizeof(t));
- for(k = 0; t[k]; ++k); //找0的位置
- r = k / 3, c = k % 3; //一维转二维
- for(int i = 0; i < 4; ++i)
- {
- nr = r + x[i], nc = c + y[i], nk = nr * 3 + nc;
- if(nr < 0 || nr > 2 || nc < 0 || nc > 2) continue;
- swap(t[k], t[nk]);
- nid = getCantor(t);
- memcpy(puz[nid], t, sizeof(t));
- swap(t[k], t[nk]);
- if(vis[nid]) continue;
- vis[nid] = 1;
- q[rear++] = nid;
- nex[nid] = id;
- dir[nid] = i;
- }
- }
- }
- int main()
- {
- char t[5], sdir[] = "durl";
- int s[M], id;
- bfs();
- while(~scanf("%s", t))
- {
- s[0] = t[0] == 'x' ? 0 : t[0] - '0';
- for(int i = 1; i < M; ++i)
- {
- scanf("%s", t);
- s[i] = t[0] == 'x' ? 0 : t[0] - '0';
- }
- id = getCantor(s);
- if(!vis[id]) puts("unsolvable");
- else
- {
- while(dir[id] >= 0)
- {
- printf("%c", sdir[dir[id]]);
- id = nex[id];
- }
- puts("");
- }
- }
- return 0;
- }
- //Last modified : 2015-07-05 11:15
Eight
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.
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
and start at the beginning of the line. Do not print a blank line between cases.
2 3 4 1 5 x 7 6 8
ullddrurdllurdruldr
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