116 lines
3.4 KiB
C++
116 lines
3.4 KiB
C++
#include <cstdio>
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#include <cstring>
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#include <unordered_set>
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#include <queue>
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#include <algorithm>
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using namespace std;
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typedef unsigned long long ULL; //自动溢出的ULL，计算整数数组的HASH值
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const int B = 17; // 17进制，用于HASH计算，这是大于15的第一个质数
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const int N = 15;
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int n;
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unordered_set<ULL> b; //用于检测一个状态是否已经访问过了
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struct Node {
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int v[N]; //当前的状态
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int step; //步数
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int f; //估值
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bool operator<(const Node &x) const { //重载小于号
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return x.f < f;
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}
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} x;
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//优先队列
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priority_queue<Node> q;
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//检测是否到了目标状态
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bool check(Node x) {
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for (int i = 0; i < n; i++)
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if (x.v[i] != i + 1) return false;
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return true;
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}
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// 计算数组状态的hash值
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ULL getHash(Node x) {
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ULL res = 0;
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for (int i = 0; i < n; i++) res = res * B + x.v[i];
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return res;
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}
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//估值函数，1次变更，可以最多修改3个后继，现在有res个后继，最少的修改次数是 ⌊res/3⌋=(res+3-1)/3
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int f(Node x) {
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int res = 0;
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for (int i = 1; i < n; i++)
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if (x.v[i] != x.v[i - 1] + 1) res++;
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return (res + 2) / 3;
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}
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// AStar
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int bfs() {
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while (q.size()) {
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Node u = q.top(); //取出当前状态
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q.pop();
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if (u.f >= 5) return 5; //当前状态无法在5步之内到达终点，返回无结果
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if (check(u)) return u.step; //如果第一次获取到了一个成功的状态，返回当前步数
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for (int len = 1; len < n; len++) { //枚举长度
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for (int st = 0; st + len - 1 < n; st++) { //枚举起始坐标
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for (int ed = st + len; ed < n; ed++) {
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//抄出来
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memcpy(x.v, u.v, sizeof u.v);
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//三次翻转，子串对调
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reverse(x.v + st, x.v + st + len);
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reverse(x.v + st + len, x.v + ed + 1);
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reverse(x.v + st, x.v + ed + 1);
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//如果此状态已经进过优先队列，那么放弃掉
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ULL hash = getHash(x);
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if (b.count(hash)) continue;
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//记录使用过
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b.insert(hash);
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//修改步数
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x.step = u.step + 1;
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//计算h值 h = g + f
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x.f = x.step + f(x);
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//入队列
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q.push(x);
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}
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}
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}
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}
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return 5;
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}
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// AC 279 ms
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int main() {
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int T;
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scanf("%d", &T);
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while (T--) {
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//清空Hash表
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b.clear();
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//清空优先队列,清空优先队列的一个小技巧
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q = {};
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scanf("%d", &n);
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//初始状态入队列
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for (int i = 0; i < n; i++) scanf("%d", &x.v[i]); //当前的数组状态
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x.step = 0; //原始状态，步数为0
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x.f = f(x); //到达终点的最少步数估值
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q.push(x); //对象入队列
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b.insert(getHash(x)); //记录此状态已使用
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int res = bfs(); // A*寻路
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if (res >= 5)
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puts("5 or more");
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else
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printf("%d\n", res);
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}
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return 0;
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}
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