P1379. 八数码难题

考点

  • 双向BFS

题解

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#include <bits/stdc++.h>
using namespace std;
// 字符串,空格的位置,到达本状态的步数
typedef pair<string, pair<int, int>> PI;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
string a, b = "123804765";

int extend(queue<PI> &aq, queue<PI> &bq, unordered_map<string, bool> &av,
unordered_map<string, bool> &bv) {
int sz = aq.size();
while (sz--) {
string str = aq.front().first;
int z = aq.front().second.first, step = aq.front().second.second;
aq.pop();
for (int i = 0; i < 4; ++i) {
int nx = (z / 3) + dx[i], ny = (z % 3) + dy[i];
if (nx < 0 || ny < 0 || nx > 2 || ny > 2) continue;
string ns = str;
int nz = nx * 3 + ny;
swap(ns[z], ns[nz]);
if (bv.count(ns)) return step + bq.front().second.second + 1;
if (av.count(ns)) continue;
av[ns] = 1;
aq.push({ns, {nz, step + 1}});
}
}
return -1;
}

int bfs() {
if (a == b) return 0;
queue<PI> aq, bq;
unordered_map<string, bool> av, bv;
for (int i = 0; i < a.length(); ++i) {
if (a[i] == '0') {
aq.push({a, {i, 0}});
break;
}
}
bq.push({b, {4, 0}});
av[a] = bv[b] = 1;
int res;
while (aq.size() && bq.size()) {
if (aq.size() < bq.size()) {
res = extend(aq, bq, av, bv);
} else {
res = extend(bq, aq, bv, av);
}
if (~res) break;
}
return res;
}

int main() {
cin >> a;
cout << bfs();
return 0;
}

思路

这道题是acwing-179. 八数码的低配版,详见该题的笔记