hdu 1430 魔板 (BFS+预处理)

　　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：

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <map>
  6 #include <stack>
  7 #include <string>
  8 
  9 using namespace std;
 10 
 11 char q[44444][10], op[44444], tmp[10];
 12 int qh, qt, last[44444];
 13 map<string, int> pos;
 14 
 15 void op1(char *s) { reverse(s, s + 8);}
 16 
 17 void op2(char *s, bool op) {
 18     if (op) {
 19         rotate(s, s + 3, s + 4);
 20         rotate(s + 4, s + 5, s + 8);
 21     } else {
 22         rotate(s, s + 1, s + 4);
 23         rotate(s + 4, s + 7, s + 8);
 24     }
 25 }
 26 
 27 void op3(char *s, bool op) {
 28     char t;
 29     if (op) {
 30         t = s[1];
 31         s[1] = s[6];
 32         s[6] = s[5];
 33         s[5] = s[2];
 34         s[2] = t;
 35     } else {
 36         t = s[1];
 37         s[1] = s[2];
 38         s[2] = s[5];
 39         s[5] = s[6];
 40         s[6] = t;
 41     }
 42 }
 43 
 44 void PRE() {
 45     for (int i = 0; i < 8; i++) tmp[i] = i + '0';
 46     tmp[8] = 0;
 47     pos.clear();
 48     qh = qt = 0;
 49 
 50     strcpy(q[qt], tmp);
 51     pos[tmp] = qt;
 52     last[qt] = -1;
 53     op[qt++] = 0;
 54     while (qh < qt) {
 55         strcpy(tmp, q[qh]);
 56         op1(tmp);
 57         if (pos.find(tmp) == pos.end()) {
 58             strcpy(q[qt], tmp);
 59             pos[tmp] = qt;
 60             last[qt] = qh;
 61             op[qt++] = 'A';
 62         }
 63 
 64         strcpy(tmp, q[qh]);
 65         op2(tmp, true);
 66         if (pos.find(tmp) == pos.end()) {
 67             strcpy(q[qt], tmp);
 68             pos[tmp] = qt;
 69             last[qt] = qh;
 70             op[qt++] = 'B';
 71         }
 72 
 73         strcpy(tmp, q[qh]);
 74         op3(tmp, true);
 75         if (pos.find(tmp) == pos.end()) {
 76             strcpy(q[qt], tmp);
 77             pos[tmp] = qt;
 78             last[qt] = qh;
 79             op[qt++] = 'C';
 80         }
 81 
 82         qh++;
 83     }
 84 }
 85 
 86 int main() {
 87 //    freopen("in", "r", stdin);
 88     PRE();
 89     char bg[10], ed[10], con[10];
 90     while (cin >> bg >> ed) {
 91         for (int i = 0; i < 8; i++) {
 92             int t = 0;
 93             while (bg[i] != ed[t]) t++;
 94             con[t] = i + '0';
 95         }
 96         con[8] = 0;
 97         int cur = pos[con];
 98         stack<char> path;
 99         while (!path.empty()) path.pop();
100         while (cur) {
101             path.push(op[cur]);
102             cur = last[cur];
103         }
104         while (!path.empty()) {
105             putchar(path.top());
106             path.pop();
107         }
108         puts("");
109     }
110     return 0;
111 }

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——written by Lyon

posted @ 2013-06-28 21:55 LyonLys 阅读(433) 评论(0) 编辑收藏举报

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