HDU 3567 BFS+预处理

HDU 1043的加强版

8数码问题

给出8数码问题的两种状态，求从A状态到B状态的最优解，数据保证有解，若有多解，输出最短且字典序最小的。

基本思路和1043的差不多，只不过这次要预处理出来9种情况的BFS

即：

BFS(0,"012345678");

BFS(1,"102345678");

BFS(2,"120345678");

BFS(3,"123045678");

BFS(4,"123405678");

BFS(5,"123450678");

BFS(6,"123456078");

BFS(7,"123456708");

BFS(8,"123456780");

然后对于每次读入的A状态根据相应的0位置，对照BFS的表进行映射，转化为1043题型的搜索。

#include "stdio.h"
#include "string.h"
#include "queue"
#include "string"
using namespace std;

const int dir[4][2]={{1,0},{0,-1},{0,1},{-1,0}};
const char indexs[]="dlru";
const int fac[]={1,1,2,6,24,120,720,5040,40320,362880};
char path[10][500010];
int time[10][500010],pri[10][500010],used[500010];

struct node
{
    int status,loc,t;
    int s[9];
};

int Cantor(int b[])
{
    int i,j,sum,cnt;
    sum=0;
    for (i=0;i<=8;i++)
    {
        cnt=0;
        for (j=i+1;j<=8;j++)
            if (b[i]>b[j])
            cnt++;
        sum+=cnt*fac[9-i-1];
    }
    return sum+1;
}
void BFS(int w,char s[])
{
    queue<node>q;
    node cur,next;
    int i,x,y,xx,yy;
    for (i=0;i<=8;i++)
        cur.s[i]=s[i]-'0';
    cur.loc=w;
    cur.t=0;
    cur.status=Cantor(cur.s);
    memset(used,0,sizeof(used));
    used[cur.status]=1;
    q.push(cur);
    pri[w][cur.status]=-1;
    while (!q.empty())
    {
        cur=q.front();
        q.pop();

        x=cur.loc/3;
        y=cur.loc%3;
        for (i=0;i<4;i++)
        {
            xx=x+dir[i][0];
            yy=y+dir[i][1];
            if (xx<0 || xx>2 || yy<0 || yy>2) continue;
            next=cur;
            next.loc=xx*3+yy;
            next.s[cur.loc]=next.s[next.loc];
            next.s[next.loc]=0;
            next.status=Cantor(next.s);
            if (used[next.status]==0)
            {
                used[next.status]=1;
                next.t++;
                path[w][next.status]=indexs[i];
                pri[w][next.status]=cur.status;
                time[w][next.status]=next.t;
                q.push(next);
            }
        }
    }
}

void output(int w,int key)
{
    if (pri[w][key]==-1) return ;
    output(w,pri[w][key]);
    printf("%c",path[w][key]);

}
int main()
{
    int T,key,j,i,Case,status;
    int a[10],b[10];
    char s1[10],s2[10];
    BFS(0,"012345678");
    BFS(1,"102345678");
    BFS(2,"120345678");
    BFS(3,"123045678");
    BFS(4,"123405678");
    BFS(5,"123450678");
    BFS(6,"123456078");
    BFS(7,"123456708");
    BFS(8,"123456780");
    scanf("%d",&T);
    Case=0;
    while (T--)
    {
        Case++;
        scanf("%s%s",s1,s2);
        if (strcmp(s1,s2)==0)
        {
            printf("Case %d: 0\n\n",Case);
            continue;
        }

        for (i=0,j=0;i<=8;i++)
        if (s1[i]=='X') { key=i; a[0]=0;}
        else a[s1[i]-'0']=++j;

        for (i=0;i<=8;i++)
        if (s2[i]=='X') b[i]=a[0];
        else b[i]=a[s2[i]-'0'];

        status=Cantor(b);

        printf("Case %d: %d\n",Case,time[key][status]);
        output(key,status);
        printf("\n");
    }
    return 0;
}