A* 算法
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#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,string> PIS;
const int N = 1e6 + 10;
string start;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
char op[] = {'u', 'r', 'd', 'l'};
int f(string s)
{
int res = 0;
for (int i = 0; i < s.size(); i ++)
if (s[i] != 'x') {
int t = s[i] - '1';
res += abs(i / 3 - t / 3) + abs(i % 3 - t % 3);
}
return res;
}
string bfs()
{
string end = "12345678x";
unordered_map<string, int> d;
unordered_map<string, pair<string,char>> pre;
priority_queue<PIS, vector<PIS>, greater<PIS>> heap;
heap.push({f(start), start});
d[start] = 0;
while (heap.size()) {
auto t = heap.top();
heap.pop();
string s = t.second;
if (s == end)
break;
int step = d[s];
int p = s.find('x');
int x = p / 3, y = p % 3;
string pre_s = s;
for (int i = 0; i < 4; i ++) {
int a = x + dx[i], b = y + dy[i];
if (a < 0 || a >= 3 || b < 0 || b >= 3)
continue;
swap(s[x * 3 + y], s[a * 3 + b]);
if (!d.count(s) || d[s] > step + 1) {
d[s] = step + 1;
pre[s] = {pre_s, op[i]};
heap.push({d[s] + f(s), s});
}
swap(s[x * 3 + y], s[a * 3 + b]);
}
}
string res;
while (end != start) {
res += pre[end].second;
end = pre[end].first;
}
reverse(res.begin(), res.end());
return res;
}
void solve()
{
char c;
while (cin >> c) {
start += c;
}
int t = 0;
for (int i = 0; i < start.size(); i ++)
for (int j = i + 1; j < start.size(); j ++) {
int s1 = start[i], s2 = start[j];
if (s1 == 'x' || s2 == 'x')
continue;
if (s1 > s2)
t ++;
}
if (t % 2)
cout << "unsolvable" << endl;
else
cout << bfs() << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
solve();
return 0;
}
- 八数码问题
设 /(s/) 为输入字符串去掉 /(x/) 后的字符串
① 当 /(s/) 的逆序对个数为奇数时,无解
② 当 /(s/) 的逆序对个数为偶数时,有解 - 估价函数
/(1/) ~ /(8/) 每个数和最终位置的曼哈顿距离之和,这个距离之和小于真实的移动次数,满足 /(A^{*}/) 算法估价值小于真实值的条件 - 在原来八数码问题的基础上,用 /(A^{*}/) 算法缩小搜索范围,并且记录方案
原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/279369.html