1150. 简单魔板

/*1150. 简单魔板初始状态1 2 3 4 8 7 6 5给定一个状态，在指定步骤内通过A、B、C操作到达目标态定义一个状态数组，头指针fp指向当前的父节点操作，rp指向子节点操作。判断是否加入子节点序列时：判重！ 扫描整个数组，若出现重复状态则跳过，对每个父节点都有三个操作，符合条件的进入队列，rp依次后移，fp完成三个操作后也fp++ */#include <stdlib.h>#include <iostream>using namespace std;struct state{  int x;  int y;  int pre;  char op;  };void OPA(int &x, int &y){   int temp = y;   y = x;   x = temp; }void OPB(int &x, int &y){   x = x%10 * 1000 + x/10;   y = y%10 * 1000 + y/10;  }void OPC(int &x, int &y){   int x1 = x/1000;   int x2 = (x - x1*1000)/100;   int x3 = (x - x1*1000 - x2*100)/10;   int x4 = x - x1*1000 - x2*100 - x3*10;      int y1 = y/1000;   int y2 = (y - y1*1000)/100;   int y3 = (y  - y1*1000 - y2*100)/10;   int y4 = y - y1*1000 - y2*100 - y3*10;      x = x1*1000 + y2*100 + x2*10 + x4;   y = y1*1000 + y3*100 + x3*10 + y4;}int main(){    int n;    int x, y;    int x1,x2,x3,x4, y1,y2,y3,y4;    state qm[81000];    char opresult[300];    n = 0;    char op[3] = {'A', 'B', 'C'};    while(cin >> n && n != -1){        cin >> x1 >> x2 >> x3 >> x4 >> y1 >> y2 >> y3 >> y4;        x = x1*1000+x2*100+x3*10+ x4;        y = y1*1000+y2*100+y3*10+ y4;        //cout << x << endl << y;        int tempx;         int tempy;        int fp=0;        int rp=1;        state init = {1234, 8765, -1, '-'};        qm[0] = init;        bool flag = true;        while(flag){                     tempx = qm[fp].x;            tempy = qm[fp].y;           //cout << tempx << " " << tempy << endl;           for(int i=0; i<3 && flag; i++){               tempx = qm[fp].x;                tempy = qm[fp].y;               if(op[i] == 'A')               {  OPA(tempx, tempy);                    //cout << op[i] << tempx << endl << tempy << endl;               }                                if(op[i] == 'B')               {  OPB(tempx, tempy);                    //cout << op[i] << tempx << endl << tempy << endl;               }                                if(op[i] == 'C')               {  OPC(tempx, tempy);                    //cout << op[i] << tempx << endl << tempy << endl;               }                             state temp =  { tempx, tempy, fp,op[i]};               if(tempx == x && tempy == y){                  flag  = false;                  qm[rp] = temp;                  rp++;               }               else {                   bool isExist = false;                   for(int k=0; k<rp; k++)                       if(qm[k].x == tempx && qm[k].y == tempy)                       {                            isExist = true;                            break;                             }                   if(!isExist){                       qm[rp] = temp;                       rp++;                            }               }               }           fp++;                // cout << fp << endl;        }       //cout << rp << endl;       int i = 0;       int temppre=rp-1;       while(qm[temppre].pre!= -1)       {           opresult[i] = qm[temppre].op;            //cout << temppre << endl;            temppre = qm[temppre].pre;                i++;            if(i>300)           {              cout << -1;              break;           }             }              if(i<=300)       {          cout << i << " " ;          for(int j=i-1; j>=0; j--)            cout << opresult[j];            }       cout << endl;    }    system("pause");       return 0;}