Aizu 0121 Seven Puzzle（從結果開始方向bfs所有可能）

題目鏈接

這題的意思是給你一個8個方格的矩形，裏面有0~7這八個數字，規定只能移動0這個方格和相鄰的方格對調，現在給你一種雜亂的情況，問你最少經過多少步才能恢復成0到7按行和列的順序排列。

挺有意思的，從來沒做過這種思路的題，就是先從結果（排好序）的情況出發，bfs遍歷所有移動0可能變成的情況，最短路徑用map<string, int>存，然後對每個詢問一一輸出就好了。

#include <bits/stdc++.h>

using namespace std;

int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
vector<int> op(20);
map<string, int> dis;
struct node {
    string s;
    int x, y, step;
    node() {}
    node(string _s, int _x, int _y, int _step) : s(_s), x(_x), y(_y), step(_step) {}
};

int input(){
    string str, ss;
    getline(cin, str);
    if (str[0] == '-')
        return -1;
    istringstream s(str);
    vector<int> v;
    v.clear();
    while(s >> ss){
        int tmp = 0;
        for (int i = 0;i < int(ss.size()); i++)
            tmp = tmp * 10 + (ss[i] - '0');
        v.push_back(tmp);
    }
    for (int i = 0;i < int(v.size()); i++)
        op[i] = v[i];
    return v.size();
}
void bfs() {
    queue<node> q;
    q.push(node("01234567", 0, 0, 0));
    dis["01234567"] = 0;
    while (!q.empty()) {
        node now = q.front();
        q.pop();
        string temp;
        for (int i = 0; i < 4; ++i) {
            int tx = now.x + dir[i][0];
            int ty = now.y + dir[i][1];
            if (tx < 0 || tx > 1 || ty < 0 || ty > 3) continue;
            temp = now.s;
            swap(temp[tx * 4 + ty], temp[now.x * 4 + now.y]);
            if (!dis[temp] && temp != "01234567") {
                dis[temp] = now.step + 1;
                q.push(node(temp, tx, ty, now.step + 1));
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.precision(10);
    cout << fixed;
#ifdef LOCAL_DEFINE
    freopen("input.txt", "r", stdin);
#endif

    bfs();
    while (input()) {
        string res = "";
        for (int i = 0; i < 8; ++i) {
            res += char(int('0') + op[i]);
        }
        cout << dis[res] << '\n';
    }

#ifdef LOCAL_DEFINE
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
    return 0;
}