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;}
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