HDU3567：Eight II(康拓展开+预处理)

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
using namespace std;

struct node
{
    int x,y;
    char map[5][5];
    node() {}
    node(char *s)
    {
        int i,j;
        int xx = 0,yy = 0;
        for(i = 0; i<strlen(s); i++)
        {
            map[xx][yy] = s[i];
            if(s[i] == 'X')
            {
                x = xx;
                y = yy;
            }
            yy++;
            if(yy == 3)
            {
                xx++;
                yy = 0;
            }
        }
    }
};

node s;
char str[20];
int num[20],hash[10];
bool vis[500000];
int pre[10][500000],ans[10][500000];
int to[4][2] = {1,0,0,-1,0,1,-1,0};
char way[10] = "dlru";

int solve(node a)//康拓
{
    int i,j,k,cnt,ans = 0;
    int b[20];
    for(i = 0; i<3; i++)
    {
        for(j = 0; j<3; j++)
        {
            b[3*i+j] = a.map[i][j];
            cnt = 0;
            for(k = 3*i+j-1; k>=0; k--)
            {
                if(b[k]>b[3*i+j])
                    cnt++;
            }
            ans+=hash[3*i+j]*cnt;
        }
    }
    return ans;
}

void bfs(int p)
{
    memset(pre[p],-1,sizeof(pre[p]));
    memset(vis,false,sizeof(vis));
    node a,next;
    queue<node> Q;
    Q.push(s);
    vis[solve(s)] = true;
    while(!Q.empty())
    {
        a = Q.front();
        Q.pop();
        int sa = solve(a);
        for(int i = 0; i<4; i++)
        {
            next = a;
            next.x+=to[i][0];
            next.y+=to[i][1];
            if(next.x<0 || next.x>2 || next.y<0 || next.y>2)
                continue;
            next.map[a.x][a.y] = next.map[next.x][next.y];
            next.map[next.x][next.y] = 'X';
            int sb = solve(next);
            if(vis[sb])
                continue;
            vis[sb] = true;
            pre[p][sb] = sa;
            ans[p][sb] = way[i];
            Q.push(next);
        }
    }
}

int main()
{
    int t,i,j,k,cas = 1;
    hash[0] = 1;
    for(i = 1; i<10; i++)
        hash[i] = hash[i-1]*i;

    s = node("X12345678");
    bfs(0);
    s = node("1X2345678");
    bfs(1);
    s = node("12X345678");
    bfs(2);
    s = node("123X45678");
    bfs(3);
    s = node("1234X5678");
    bfs(4);
    s = node("12345X678");
    bfs(5);
    s = node("123456X78");
    bfs(6);
    s = node("1234567X8");
    bfs(7);
    s = node("12345678X");
    bfs(8);

    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",str);
        int p;
        for(i = 0,j = 0; i<9; i++)//保存位置，因为前面预处理的都是位置
        {
            if(str[i]=='X') p = i;
            else
                num[str[i]-'0'] = j++;
        }
        scanf("%s",str);
        for(i = 0; i<9; i++)//求出目标状态每个数在原状态的位置
        {
            if(str[i]=='X')
                continue;
            str[i] = num[str[i]-'0']+'1';
        }
        s = node(str);//由目标态逆推到初始态
        int sum = solve(s);
        string ss="";
        while(sum!=-1)
        {
            ss+=ans[p][sum];
            sum = pre[p][sum];
        }
        printf("Case %d: %d\n",cas++,ss.size()-1);
        for(i = ss.size()-2; i>=0; i--)//由于方案是逆推，输出也要逆推
            printf("%c",ss[i]);
        printf("\n");
    }

    return 0;
}