题意不说了，就是一个八数码
思路：
很暴力，直接把空格看成0，然后把九个数字连接起来，用这个数来代表这个状态。
然后正常来想这个题目时，肯定是 从给定状态到目标态来bfs搜索，但是这样做死活爆内存 （好难受= =）
我们需要转个弯，从目标态向给定态搜索，这样可以直接预处理打表了。
正好这是一个9个不同数字的排列，用康托定理完美去重。
注： 用c++交的，G++还是爆内存= =
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int md = 370000 + 7;
char cmd[7];
queue<int>q;
char tab[5][5];
string state[md];
const int dx[] = {0,0,-1,1}; /// 0 - l, 1 - r, 2 - u, 3 - d;
const int dy[] = {-1,1,0,0};
const char* flag = "rldu";
int jie[10];
void init(){
jie[0] = jie[1] = 1;
for (int i = 2; i < 9; ++i){
jie[i] = jie[i-1] * i;
}
}
char hashs[10];
int Hash(int v){
for (int i = 0; i < 9; ++i){
hashs[9-i-1] = v % 10 + 48;
v /= 10;
}
int ans = 0;
for (int i = 0; i < 9; ++i){
int sum = 0;
for (int j = i+1; j < 9; ++j){
if (hashs[j] < hashs[i])++sum;
}
ans += sum * jie[9-i-1];
}
return ans;
}
bool vis[md];
void bfs(int vvvv){
while(!q.empty()) q.pop();
q.push(vvvv);
int v = 0;
string tu = "";
vis[Hash(vvvv)] = 1;
while(!q.empty()){
int u = q.front(); q.pop();
int idu = Hash(u);
tu = "";
for (int i = 0; i < 9; ++i){
tu += u % 10 + 48;
u/=10;
}
reverse(tu.begin(),tu.end());
int pos;
for (pos=0;pos<9; ++pos) if (tu[pos] =='0')break;
for (int i = 0; i < 9; ++i){
tab[i/3][i%3] = tu[i];
}
int x = pos/3; int y = pos % 3;
for (int i = 0 ; i < 4; ++i){
int xx = dx[i] + x;
int yy = dy[i] + y;
if (xx >= 0 && xx <= 2 && yy >=0 && yy <= 2){
swap(tab[x][y],tab[xx][yy]);
v = 0;
for (int j = 0; j < 3; ++j){
for (int k = 0; k < 3; ++k){
v = v * 10 + tab[j][k] - 48;
}
}
int idv = Hash(v);
if (!vis[idv]){
vis[idv] = 1;
state[idv] = state[idu] + flag[i];
q.push(v);
}
swap(tab[x][y],tab[xx][yy]);
}
}
}
}
int main(){
init();
// printf("%d\n",Hash(876543210));
memset(vis,0,sizeof vis);
for (int i = 0; i < md; ++i){
state[i] = "";
}
bfs(123456780);
for (int i = 0; i < md; ++i){
reverse(state[i].begin(),state[i].end());
}
while(~scanf("%s",cmd)){
if (cmd[0] == 'x')cmd[0] = 48;
int v = cmd[0] - 48;
for (int i = 1; i < 9; ++i){
scanf("%s",cmd);
if (cmd[0] == 'x') cmd[0] = '0';
v = v * 10 + cmd[0] - 48;
}
if (!vis[Hash(v)]) puts("unsolvable");
else printf("%s\n",state[Hash(v)].c_str());
}
return 0;
}
Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22065 Accepted Submission(s): 5922
Special Judge
Problem Description
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.
Input
You will receive, several descriptions of configuration of the 8 puzzle. One 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
Output
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. Do not print a blank line between cases.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
Source
South Central USA 1998 (Sepcial Judge Module By JGShining)
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