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HDU 1043 Eight 八数码
类别:C/C++编程   作者:码皇   来源:star_sky hht     点击:

EightTime Limit : 10000 5000ms (Java Other) Memory Limit : 65536 32768K (Java Other)Total Submission(s) : 15 Accepted Submission(s) : 2Special JudgeProblem Description The

 

Eight

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 15 Accepted Submission(s) : 2
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 1213 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 1213 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)

 

 

第一次接触A* 写得太想吐了。。。

 

 

    #include#include#include#include#include#include#include#include#include#define inf 1<<30#define eps 1e-7#define LD long double#define LL long long#define maxn 1000000005using namespace std;
    const int ed=322560;
    int Hash[9]= {
    1,1,2,6,24,120,720,5040,40320}
    ;
    struct node{
    int Map[3][3];
    int h,g;
    int x,y;
    int Hash;
    bool operator<(const node n1)const //优先队列第一关键字为h,第二关键字为g {
    return h!=n1.h?h>n1.h:g>n1.g;
    }
    bool cheak() {
    if(x>=0&&x<3&&y>=0&&y<3) return true ;
    return false ;
    }
    }
    u,v;
    int vis[400000];
    int pri[400000];
    int dir[4][2]= {
    0,1,0,-1,1,0,-1,0}
    ;
    bool ok(node tmp) //判断状态是否合法 逆序数为偶数{
    int a[9],k=0;
    for(int i=0;
    i<3;
    i++) for(int j=0;
    j<3;
    j++) a[k++]=tmp.Map[i][j];
    int ans=0;
    for(int i=0;
    i int="" j="i+1;
    ">a[j]) ans++;
    return (ans&1);
    }
    int get_hash(node tmp) //获得康拓值{
    int ans=0;
    int a[9],k=0;
    for(int i=0;
    i<3;
    i++) for(int j=0;
    j<3;
    j++) a[k++]=tmp.Map[i][j];
    for(int i=0;
    i int="" j="0;
    " sum="0;
    ">a[i]) sum++;
    ans+=Hash[i]*sum;
    }
    return ans;
    }
    int get_h(node tmp) //估价函数 每一个点移动到原来地方的最短路程和 曼哈顿距离{
    int ans;
    for(int i=0;
    i<3;
    i++) for(int j=0;
    j<3;
    j++) if(tmp.Map[i][j]) ans+=abs(i-(tmp.Map[i][j]-1)/3)+abs(j-(tmp.Map[i][j]-1)%3);
    return ans;
    }
    void print(){
    string str;
    str.clear();
    int st=ed;
    while(pri[st]!=-1) {
    if(vis[st]==0) {
    str+='
    r'
    ;
    }
    else if(vis[st]==1) {
    str+='
    l'
    ;
    }
    else if(vis[st]==2) {
    str+='
    d'
    ;
    }
    else str+='
    u'
    ;
    st=pri[st];
    }
    for(int i=str.size()-1;
    i>=0;
    i--) cout< node="" return="" void="">q;
    q.push(u);
    memset(vis,-1,sizeof(vis));
    memset(pri,-1,sizeof(pri));
    vis[u.Hash]=-2;
    while(!q.empty()) {
    //cout< char="" cin="" i="0;
    " int="" return="" u="q.top();
    " v="u;
    " v.h="get_h(v);
    " v.hash="=ed)">>ch[0]) {
    for(int i=1;
    i<9;
    i++) cin>>ch[i];
    for(int i=0;
    i<3;
    i++) {
    for(int j=0;
    j<3;
    j++) {
    if(ch[i*3+j]=='
    x'
    ) {
    u.Map[i][j]=0;
    u.x=i;
    u.y=j;
    u.Hash=get_hash(u);
    u.h=get_h(u);
    u.g=0;
    }
    else u.Map[i][j]=ch[i*3+j]-'
    0'
    ;
    }
    }
    if(u.Hash==ed) {
    cout< continue="" else="" pre="">

     

     

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