95 lines
2.8 KiB
C++
95 lines
2.8 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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typedef unordered_map<string, int> MSI;
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typedef unordered_map<string, pair<char, string>> MSP;
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string st, ed = "12345678x";
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// 上下左右
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char op[] = {'u', 'd', 'l', 'r'};
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int dx[] = {-1, 1, 0, 0};
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int dy[] = {0, 0, -1, 1};
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struct Node {
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int dist;
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string str;
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const bool operator<(const Node &b) const {
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return b.dist < dist;
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}
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};
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// 前驱
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MSP pre;
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MSI dist;
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priority_queue<Node> q;
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// 估值函数【每个位置的现实值与理想值的曼哈顿距离差值和】
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int f(string str) {
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int res = 0;
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for (int i = 0; i < str.size(); i++)
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if (str[i] != 'x') {
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int t = str[i] - '1'; // str[i]对应的数字，应该在下标从0开始的字符串哪个位置
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// ① 利用除3模3大法，将下标从0开始的坐标转换为二维坐标,并计算理想值与目标值的曼哈顿距离
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// ② 曼哈顿距离的累加值才是当前状态与理想状态的整体差距
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res += abs(i / 3 - t / 3) + abs(i % 3 - t % 3);
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}
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return res;
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}
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string astar() {
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// 将估价函数与字符串形成数对，放入到优先队列中
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q.push({f(st), st});
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dist[st] = 0; // 记录距离为0
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while (q.size()) {
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string t = q.top().str;
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q.pop();
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if (t == ed) break;
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int k = t.find('x');
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int tx = k / 3, ty = k % 3;
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for (int i = 0; i < 4; i++) { // 枚举udlr
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int x = tx + dx[i], y = ty + dy[i];
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if (x < 0 || x >= 3 || y < 0 || y >= 3) continue; // 出界
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string v = t; // 将t字符串复制出来一个，生成 v字符串
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swap(v[k], v[x * 3 + y]); // 将原字符串中x与目标位置进行交换，生成新的目标状态字符串v
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if (dist[v]) continue;
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dist[v] = dist[t] + 1; // 更新最小值
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q.push({dist[v] + f(v), v}); // 将新的估价函数计算更新，重新入队列
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pre[v] = {op[i], t};
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}
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}
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// 将路径倒转过来
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string res;
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while (ed != st) {
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res += pre[ed].first;
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ed = pre[ed].second;
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}
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reverse(res.begin(), res.end());
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return res;
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}
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int main() {
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char c;
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for (int i = 0; i < 9; i++) cin >> c, st += c;
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// 八数码定理
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int nums = 0;
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for (int i = 0; i < 9; i++) {
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if (st[i] == 'x') continue;
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for (int j = i + 1; j < 9; j++) {
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if (st[j] == 'x') continue;
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if (st[j] < st[i]) nums++; // 逆序数
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}
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}
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// 奇数个逆序对，无解
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if (nums & 1)
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puts("unsolvable");
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else
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cout << astar() << endl; // 运行astar算法，返回最短距离
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return 0;
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}
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