hdu 1430 魔板 (BFS+预处理)

Problem - 1430

　　跟八数码相似的一题搜索题。做法可以是双向BFS或者预处理从"12345678"开始可以到达的所有状态，然后等价转换过去直接回溯路径即可。

代码如下：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <string>

 using namespace std;

 char q[][], op[], tmp[];
 int qh, qt, last[];
 map<string, int> pos;

 void op1(char *s) { reverse(s, s + );}

 void op2(char *s, bool op) {
     if (op) {
         rotate(s, s + , s + );
         rotate(s + , s + , s + );
     } else {
         rotate(s, s + , s + );
         rotate(s + , s + , s + );
     }
 }

 void op3(char *s, bool op) {
     char t;
     if (op) {
         t = s[];
         s[] = s[];
         s[] = s[];
         s[] = s[];
         s[] = t;
     } else {
         t = s[];
         s[] = s[];
         s[] = s[];
         s[] = s[];
         s[] = t;
     }
 }

 void PRE() {
     for (int i = ; i < ; i++) tmp[i] = i + '';
     tmp[] = ;
     pos.clear();
     qh = qt = ;

     strcpy(q[qt], tmp);
     pos[tmp] = qt;
     last[qt] = -;
     op[qt++] = ;
     while (qh < qt) {
         strcpy(tmp, q[qh]);
         op1(tmp);
         if (pos.find(tmp) == pos.end()) {
             strcpy(q[qt], tmp);
             pos[tmp] = qt;
             last[qt] = qh;
             op[qt++] = 'A';
         }

         strcpy(tmp, q[qh]);
         op2(tmp, true);
         if (pos.find(tmp) == pos.end()) {
             strcpy(q[qt], tmp);
             pos[tmp] = qt;
             last[qt] = qh;
             op[qt++] = 'B';
         }

         strcpy(tmp, q[qh]);
         op3(tmp, true);
         if (pos.find(tmp) == pos.end()) {
             strcpy(q[qt], tmp);
             pos[tmp] = qt;
             last[qt] = qh;
             op[qt++] = 'C';
         }

         qh++;
     }
 }

 int main() {
 //    freopen("in", "r", stdin);
     PRE();
     char bg[], ed[], con[];
     while (cin >> bg >> ed) {
         for (int i = ; i < ; i++) {
             int t = ;
             while (bg[i] != ed[t]) t++;
             con[t] = i + '';
         }
         con[] = ;
         int cur = pos[con];
         stack<char> path;
         while (!path.empty()) path.pop();
         while (cur) {
             path.push(op[cur]);
             cur = last[cur];
         }
         while (!path.empty()) {
             putchar(path.top());
             path.pop();
         }
         puts("");
     }
     return ;
 }

——written by Lyon