HDU-1043 Eight（A*）

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15894    Accepted Submission(s): 4386
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
   
   
   
   

今年在北大暑期学校听讲时，郭老师讲过，但他的代码不能完全理解，特别是排列和序号相互转换。

此次再遇八数码问题，看到别的大神用康托展开计算序号，感觉十分方便，应该牢记。

但看到大神代码，优先队列以h为第一关键字，g为第二关键字能AC且450ms；

我自己本地跑，连样例都无法过，不过以f为第一关键字，h为第二关键字的方法比只以f为关键字的方法快大概200ms

#include <cstdio>
#include <cstring>
#include <string>
#include <queue>

using namespace std;

struct Node {
    int a[9],pos,has,h,g;
    bool operator < (const Node& a) const {
        if(h+g!=a.h+a.g)
            return h+g>a.h+a.g;
        return h!=a.h?h>a.h:g>a.g;
    }
}sta,u,v;

inline bool up(const Node& t) {
    return t.pos>2;
}
inline bool right(const Node& t) {
    return t.pos!=2&&t.pos!=5&&t.pos!=8;
}
inline bool down(const Node& t) {
    return t.pos<6;
    }
inline bool left(const Node& t) {
    return t.pos!=0&&t.pos!=3&&t.pos!=6;
}
bool (*tab[4])(const Node& t)={up,right,down,left};

int HASH[9]={40320,5040,720,120,24,6,2,1,1};//n的阶乘，变进制
int turn[400001];
int pre[400001];
const int des=46233;
const int d[]={-3,1,3,-1};
const char m[]={"urdl"};
const int hs[9][9]={0,0,1,2,1,2,3,2,3,//位置hs[i][j]表示位置i数字为j时与正确位置的曼哈顿距离
                    0,1,0,1,2,1,2,3,2,
                    0,2,1,0,3,2,1,4,3,
                    0,1,2,3,0,1,2,1,2,
                    0,2,1,2,1,0,1,2,1,
                    0,3,2,1,2,1,0,3,2,
                    0,2,3,4,1,2,3,0,1,
                    0,3,2,3,2,1,2,1,0,
                    0,4,3,2,3,2,1,2,1};

bool judge() {//奇偶剪枝
    int num=0,i,j;
    for(i=0;i<9;++i)
        for(j=i+1;j<9;++j)
            if(sta.a[i]&&sta.a[j]&&sta.a[i]>sta.a[j])
                ++num;
    return (num&1)==1;
}

int get_hash(const Node& t) {
    int num=0,i,j,k;
    for(i=0;i<9;++i) {
        k=0;
        for(j=i+1;j<9;++j)
            if(t.a[i]>t.a[j])
                ++k;
        num+=HASH[i]*k;
    }
    return num;
}

int get_h(const Node& t) {//计算估价函数h，曼哈顿距离和
    int num=0;
    for(int i=0;i<9;++i)
        num+=hs[i][t.a[i]];
    return num;
}

void astar() {
    int i;
    priority_queue<Node> q;
    q.push(sta);
    while(!q.empty()) {
        u=q.top();
        q.pop();
        for(i=0;i<4;++i) {
            if(tab[i](u)) {
                v=u;
                v.pos+=d[i];
                swap(v.a[u.pos],v.a[v.pos]);
                if(turn[v.has=get_hash(v)]==-1) {
                    turn[v.has]=i;
                    ++v.g;
                    v.h=get_h(v);
                    pre[v.has]=u.has;
                    q.push(v);
                }
                if(v.has==des)
                    return ;
            }
        }
    }
}

void print() {
    string ans;
    int nxt=des;
    while(pre[nxt]!=-2) {
        ans.push_back(m[turn[nxt]]);
        nxt=pre[nxt];
    }
    for(int i=ans.size()-1;i>=0;--i)
        printf("%c",ans[i]);
    printf("\n");
}

int main() {
    //freopen("out.txt","w",stdout);
    int i;
    char s[3];
    while(scanf("%s",s)==1) {
        if(s[0]=='x') {
            sta.a[0]=0;
            sta.pos=0;
        }
        else
            sta.a[0]=s[0]-'0';
        for(i=1;i<9;++i) {
            scanf("%s",s);
            if(s[0]=='x') {
                sta.a[i]=0;
                sta.pos=i;
            }
            else
                sta.a[i]=s[0]-'0';
        }
        if(judge()) {
            printf("unsolvable\n");
            continue;
        }
        if((sta.has=get_hash(sta))==des) {
            printf("\n");
            continue;
        }
        memset(turn,-1,sizeof(turn));
        memset(pre,-1,sizeof(pre));
        pre[sta.has]=-2;
        turn[sta.has]=0;
        sta.g=0,sta.h=get_h(sta);
        astar();
        print();
    }
    return 0;
}