tjut 3567

来源：互联网发布：centos rm 恢复编辑：程序博客网时间：2024/06/10 01:57
#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;  }
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