POJ 1077 A*

链接：

http://poj.org/problem?id=1077

题意：

经典8数码问题，直接暴力bfs也能做，但是一定要先hash一下

题解：

这里的估价函数为当前状态下，所有的数字与其位置的之差的绝对值总和

话说我又被c++的string坑惨了

代码：

 31 int vis[MAXN];
 32 int fac[] = { 1,1,2,6,24,120,720,5040,40320,362880 };
 33
 34 int find(char s[]) {//康托展开
 35     int res = 0;
 36     bool has[10] = { 0 };
 37     rep(i, 0, 9) {
 38         int x = s[i] - ‘0‘, y = 0;
 39         rep(j, 1, x) if (!has[j]) y++;
 40         res += y*fac[8 - i];
 41         has[x] = 1;
 42     }
 43     return res;
 44 }
 45
 46 int get_h(const char s[]) {
 47     int res = 0;
 48     rep(i, 0, 9) res += abs(s[i] - ‘1‘ - i);
 49     return res;
 50 }
 51
 52 struct Node {
 53     char s[15];
 54     int g;
 55     Node(char s[], int g) :g(g) { strcpy(this->s, s); }
 56     const bool operator<(const Node &t) const {
 57         return g + get_h(s) > t.g + get_h(t.s);
 58     }
 59 };
 60
 61 priority_queue<Node> que;
 62
 63 int main() {
 64     char s[15];
 65     rep(i, 0, 9) {
 66         string t;
 67         cin >> t;
 68         if (t[0] == ‘x‘) s[i] = ‘9‘;
 69         else s[i] = t[0];
 70     }
 71     vis[find(s)] = 100;
 72     que.push(Node(s, 0));
 73     while (!que.empty()) {
 74         Node pre = que.top(); que.pop();
 75         int x = 0;
 76         while (pre.s[x] != ‘9‘) x++;
 77         if (x >= 3) {
 78             char ss[15];
 79             strcpy(ss, pre.s);
 80             ss[x] = ss[x - 3];
 81             ss[x - 3] = ‘9‘;
 82             if (!vis[find(ss)]) {
 83                 vis[find(ss)] = -3;
 84                 que.push(Node(ss, pre.g + 1));
 85             }
 86         }
 87         if (x < 6) {
 88             char ss[15];
 89             strcpy(ss, pre.s);
 90             ss[x] = ss[x + 3];
 91             ss[x + 3] = ‘9‘;
 92             if (!vis[find(ss)]) {
 93                 vis[find(ss)] = 3;
 94                 que.push(Node(ss, pre.g + 1));
 95             }
 96         }
 97         if (x % 3) {
 98             char ss[15];
 99             strcpy(ss, pre.s);
100             ss[x] = ss[x - 1];
101             ss[x - 1] = ‘9‘;
102             if (!vis[find(ss)]) {
103                 vis[find(ss)] = -1;
104                 que.push(Node(ss, pre.g + 1));
105             }
106         }
107         if (x % 3 != 2) {
108             char ss[15];
109             strcpy(ss, pre.s);
110             ss[x] = ss[x + 1];
111             ss[x + 1] = ‘9‘;
112             if (!vis[find(ss)]) {
113                 vis[find(ss)] = 1;
114                 que.push(Node(ss, pre.g + 1));
115             }
116         }
117         if (vis[0]) break;
118     }
119     if (!vis[0]) return puts("unsolvable"), 0;
120     int y = 0;
121     char ss[15] = "123456789";
122     string ans;
123     while (vis[y] != 100) {
124         int x = 0;
125         while (ss[x] != ‘9‘) x++;
126         switch (vis[y]) {
127         case -3: {
128             ss[x] = ss[x + 3];
129             ss[x + 3] = ‘9‘;
130             ans += ‘u‘;
131             break;
132         }
133         case 3: {
134             ss[x] = ss[x - 3];
135             ss[x - 3] = ‘9‘;
136             ans += ‘d‘;
137             break;
138         }
139
140         case -1: {
141             ss[x] = ss[x + 1];
142             ss[x + 1] = ‘9‘;
143             ans += ‘l‘;
144             break;
145         }
146         case 1: {
147             ss[x] = ss[x - 1];
148             ss[x - 1] = ‘9‘;
149             ans += ‘r‘;
150             break;
151         }
152         }
153         y = find(ss);
154     }
155     reverse(all(ans));
156     cout << ans << endl;
157     return 0;
158 }