Eight II HDU - 3567 （bfs打表+hash映射+康托展开）
Eight II
HDU - 3567
Eight-puzzle, which is also called "Nine grids", comes from an old game.
In this game, you are given a 3 by 3 board and 8 tiles. The tiles are numbered from 1 to 8 and each covers a grid. As you see, there is a blank grid which can be represented as an 'X'. Tiles in grids having a common edge with the blank grid can be moved into that blank grid. This operation leads to an exchange of 'X' with one tile.
We use the symbol 'r' to represent exchanging 'X' with the tile on its right side, and 'l' for the left side, 'u' for the one above it, 'd' for the one below it.
A state of the board can be represented by a string S using the rule showed below.
The problem is to operate an operation list of 'r', 'u', 'l', 'd' to turn the state of the board from state A to state B. You are required to find the result which meets the following constrains:
1. It is of minimum length among all possible solutions.
2. It is the lexicographically smallest one of all solutions of minimum length.
In this game, you are given a 3 by 3 board and 8 tiles. The tiles are numbered from 1 to 8 and each covers a grid. As you see, there is a blank grid which can be represented as an 'X'. Tiles in grids having a common edge with the blank grid can be moved into that blank grid. This operation leads to an exchange of 'X' with one tile.
We use the symbol 'r' to represent exchanging 'X' with the tile on its right side, and 'l' for the left side, 'u' for the one above it, 'd' for the one below it.
A state of the board can be represented by a string S using the rule showed below.
The problem is to operate an operation list of 'r', 'u', 'l', 'd' to turn the state of the board from state A to state B. You are required to find the result which meets the following constrains:
1. It is of minimum length among all possible solutions.
2. It is the lexicographically smallest one of all solutions of minimum length.
Input The first line is T (T <= 200), which means the number of test cases of this problem.
The input of each test case consists of two lines with state A occupying the first line and state B on the second line.
It is guaranteed that there is an available solution from state A to B.
Output For each test case two lines are expected.
The first line is in the format of "Case x: d", in which x is the case number counted from one, d is the minimum length of operation list you need to turn A to B.
S is the operation list meeting the constraints and it should be showed on the second line.
Sample Input
2 12X453786 12345678X 564178X23 7568X4123Sample Output
Case 1: 2 dd Case 2: 8 urrulldr
这道题是上一道题Eight I的一道升级版，这道题首先需要从指定的初始状态到指定的目标状态，但是我们可以发现的是总的大情况无外乎就一下九种
char eight_num[9][10] = {"X12345678" , "1X2345678" , "12X345678" , "123X45678" ,
"1234X5678" , "12345X678" , "123456X78" , "1234567X8" , "12345678X" }; 12X453786我们就可以做这样的一个映射“12X345678”
分别是：1 → 1，2 → 2，X → X，4 → 3，5 → 4，3 → 5，7 → 6，8 → 7，6 → 8，
然后目标状态一样要映射掉。
这样我们就可以通过预处理，将以每种状态为起点状态开始bfs存下它所能到达的所有状态的路径，但是这个题还有个特殊要求要求长度最短，字典序最小，那么记录路径就不能想上个题一样直接方向和字符相反对应了。
为了使得字典序最小我们必须要让方向以dlru（down，left，right，up）的顺序进行bfs（字母顺序），这样我们预处理完记录下路径后，我们只能从得到的目标状态往前找到初始状态，然后在逆序输出。
具体代码如下：
#include
#include
#include
#include
#include
using namespace std;
const int INF = 1e8;
const int MAXN = 4e5;
int n,m;
char eight_num[9][10] = {"X12345678","1X2345678","12X345678","123X45678","1234X5678","12345X678",
"123456X78","1234567X8","12345678X"};
char step[] = "dlru";
int dx[4] = {1,0,0,-1};
int dy[4] = {0,-1,1,0};
int fac[9] = {1,1,2,6,24,120,720,5040,40320};
int vis[9][MAXN];
int pre[9][MAXN];
struct node{
int num[9];//八数码
int state;//状态（康托展开值）
int pos;//x的位置
};
//得到状态
int Cantor(int *s){
int sum = 0;
for(int i = 0; i < 9; i++){
int num = 0;
for(int j = i+1; j < 9; j++){
if(s[i] > s[j]) num++;
}
sum += num * fac[8-i];
}
return sum;
}
void printAns(int state,int kind){
string ans;
while(state != -1){
ans += step[vis[kind][state]];
state = pre[kind][state];
}
printf("%d\n",ans.size()-1);
for(int i = ans.size() - 2; i >= 0; i--){
printf("%c",ans[i]);
}
puts("");
}
void bfs(node x,int kind){
queue q;
q.push(x);
vis[kind][x.state] = 1;//这个是初始化的值并不是真正走的值所以上面输出的时候是从倒数第二个开始输出
while(!q.empty()){
node u = q.front();
q.pop();
//4种交换
int x = u.pos / 3;
int y = u.pos % 3;
for(int i = 0; i < 4; i++){
int xx = x + dx[i];
int yy = y + dy[i];
if(xx >= 0 && xx < 3 && yy >= 0 && yy < 3){
node v = u;
swap(v.num[u.pos],v.num[xx*3+yy]);
v.state = Cantor(v.num);
if(vis[kind][v.state] == -1){
v.pos = xx * 3 + yy;
vis[kind][v.state] = i;//代表通过i方向的移动到达kind这类中的v.state状态
pre[kind][v.state] = u.state;//记录这种状态的前一种状态
q.push(v);
}
}
}
}
}
void init(char *s,int kind){
node u;
for(int i = 0; i < 9; i++){
if(s[i] == 'X'){
u.num[i] = 0;
u.pos = i;
}
else u.num[i] = s[i] - '0';
}
u.state = Cantor(u.num);
bfs(u,kind);
}
int main(){
int t,cas = 0;
char s[15],c[15];
int kind;
int temp[MAXN];
memset(vis,-1,sizeof(vis));
memset(pre,-1,sizeof(pre));
for(int i = 0; i < 9; i++){
init(eight_num[i],i);
}
scanf("%d",&t);
while(t--){
scanf("%s",s);
int cnt = 1;
//映射初始状态的八数码
for(int i = 0; i < 9; i++){
if(s[i] == 'X'){
c[0] = 0;
kind = i;
}
else c[s[i]-'0'] = cnt++;
}
//映射目标状态八数码
scanf("%s",s);
for(int i = 0; i < 9; i++){
if(s[i] == 'X') temp[i] = c[0];
else temp[i] = c[s[i]-'0'];
}
int state = Cantor(temp);
printf("Case %d: ",++cas);
printAns(state,kind);
}
return 0;
}