Poj 1077 Eight！（BFS-A*）

1077 -- Eight (poj.org)

由结论可以知道逆序对为奇数的时候无解，f为估值函数，计算曼哈顿距离。

想用康托展开写，但多状态数码问题用数组存储状态不够，有TLE的风险，还是A*吧！

吃一个tomato宣告今日...不知道结不结束

#include <iostream>
#include <vector>
#include <queue>
#include <unordered_map>
using namespace std;
const int dx[] = {-1,0,1,0},dy[] = {0,1,0,-1};
const char op[] = "urdl";
typedef pair <int,string> pii;
char ch;
string start,s;
int f (string state) {
    int ans = 0;
    for (int i = 0;i < 9;i++) {
        if (state[i] == 'x') continue;
        int t = state[i] - '1';
        ans += abs (i / 3 - t / 3) + abs (i % 3 - t % 3);
    }
    return ans;
}
string Astar () {
    string end = "12345678x";
    unordered_map <string,int> dist;
    unordered_map <string,pair <char,string>> prev;
    priority_queue <pii,vector <pii>,greater <pii>> heap;
    dist[start] = 0;
    heap.push ({f(start),start});
    while (!heap.empty ()) {
        auto t = heap.top ();
        heap.pop ();
        string state = t.second;
        if (state == end) break;
        int x,y;
        for (int i = 0;i < 9;i++) {
            if (state[i] == 'x') {
                x = i / 3,y = i % 3;
                break;
            }
        }
        string source = state;
        for (int i = 0;i < 4;i++) {
            int a = x + dx[i],b = y + dy[i];
            if (a < 0 || a > 2 || b < 0 || b > 2) continue;
            swap (state[a*3+b],state[x*3+y]);
            if (!dist.count (state) || dist[state] > dist[source] + 1) {
                dist[state] = dist[source] + 1;
                prev[state] = {op[i],source};
                heap.push ({dist[state] + f(state),state});
            }
            state = source;
        }
    }
    string ans;
    while (end != start) {
        ans = prev[end].first + ans;
        end = prev[end].second;
    }
    return ans;
}
signed main () {
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    for (int i = 0;i < 9;i++) {
        cin >> ch;
        start += ch;
        if (ch != 'x') s += ch;
    }
    int cnt = 0;
    for (int i = 0;i < 8;i++) {
        for (int j = i;j < 8;j++) {
            if (s[i] > s[j]) cnt++;
        }
    }
    if (cnt & 1) puts ("unsolvable");
    else cout << Astar () << endl;
    return 0;
}