83 lines
2.3 KiB
C++
83 lines
2.3 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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string st, ed = "12345678x";
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//上右下左
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int dx[] = {-1, 0, 1, 0};
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int dy[] = {0, 1, 0, -1};
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//伟大的结构体放光芒
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struct Node {
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int dist;
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string str;
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bool operator<(const Node &t) const {
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return dist > t.dist; //距离大的反而小，距离小的反而大。所以，配合优先队列（大顶堆），小的在上，大的在下
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}
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};
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//估值函数,用曼哈顿距离估值
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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';
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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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int astar() {
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MSI dist;
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priority_queue<Node> q;
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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++) {
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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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}
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}
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//返回结果
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return dist[ed];
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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("-1");
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else //运行astar算法，返回最短距离
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printf("%d\n", astar());
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return 0;
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}
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