POJ 1077-Eight(BFS+优先队列)

[align=center]Eight[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28463		Accepted: 12420		Special Judge

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 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

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.
Sample Input

2  3  4  1  5  x  7  6  8

Sample Output

ullddrurdllurdruldr

Source
South Central USA 1998

题意：
  给出一个八数码给你，要你求出1，2，3，4，5，6，7，8，x的顺序。

思路：
  这题很明显用BFS来做，当我写好了代码才发现我竟然不会如何去剪枝，所以一直跑重复的点。之后去看了看题解发现直接用hash去保存九个位置就是这个算法的剪枝了，思维短路了，竟然没意识到应该用hash去剪枝。

AC代码：

#include<iostream>
#include<functional>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<queue>
#include<cmath>
#include<map>
#include<set>
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
#define QWQ ios::sync_with_stdio(0)
#define inf 0x3f3f3f3f
typedef unsigned long long LL;
typedef  long long ll;

const int T = 500000+50;
const int mod = 1000000007;
char s[5][5];
int xy[][2]={{1,0},{0,-1},{-1,0},{0,1}};
char fx[] = {{'r'},{'u'},{'l'},{'d'}};

struct node
{
char s[5][5];
vector<char> ve;
int c,x,y;
node(){}
bool operator<(const node& b)const{
return c<b.c;
}
};

set<int> se;

inline int jugde(const node& b)
{
int k=0;
for(int i = 0;i<3;++i){
for(int j=0;j<3;++j){
if(b.s[i][j]=='0'+i*3+j+1)k++;
}
}
return k;
}
inline void Copy1(node& a,const node& b)
{
for(int i=0;i<3;++i){
a.s[i][0]=b.s[i][0],
a.s[i][1]=b.s[i][1],
a.s[i][2]=b.s[i][2];
a.s[i][3]='\0';
}
}

int Hash(const node& a)
{
int sum = 0;
for(int i=0;i<3;++i){
for(int j=0;j<3;++j){
if(a.s[i][j]=='x'){
sum = sum*10;
}
else {
sum = sum*10 + a.s[i][j]-'0';
}
}
}
return sum;
}

void BFS(int x,int y)
{
se.clear();
priority_queue<node> q;
node t,tt;

for(int i=0;i<3;++i){
t.s[i][0]=s[i][0],
t.s[i][1]=s[i][1],
t.s[i][2]=s[i][2];
t.s[i][3]='\0';
}
t.x=x,t.y=y;
t.c = jugde(t);
q.push(t);
se.insert(Hash(t));
while(!q.empty())
{
t = q.top();q.pop();
if(t.c==8){
for(int i=0;i<t.ve.size();++i){
printf("%c",t.ve[i]);
}
printf("\n");
break;
}

for(int j=0;j<4;++j){
int tx = t.x + xy[j][1];
int ty = t.y + xy[j][0];
if(tx>=0&&tx<3&&ty>=0&&ty<3){
tt.x = tx;
tt.y = ty;

Copy1(tt,t);
swap(tt.s[tx][ty],tt.s[t.x][t.y]);
tt.c = jugde(tt);
tt.ve = t.ve;
tt.ve.push_back(fx[j]);
int tmp = Hash(tt);
if(se.find(tmp)==se.end()){
se.insert(tmp);
q.push(tt);
}
}
}
}
}

int main()
{

#ifdef zsc
freopen("input.txt","r",stdin);
#endif

int i,j,u,v;
while(~scanf(" %c",&s[0][0]))
{
for(i=1;i<9;++i){
scanf(" %c",&s[i/3][i%3]);
if(s[i/3][i%3]=='x')u=i/3,v=i%3;
}
BFS(u,v);
}

return 0;
}