洛谷 P1379 八数码难题

题意

八数码问题

2025-6-2更新:
学习了Astar算法 , 加上了Astar的代码

代码

#include<bits/stdc++.h>
using namespace std;
const int es = 123804765;
int dx[] = {-1,0,1,0};
int dy[] = {0,-1,0,1};
unordered_map<int,int>mp;
unordered_map<int,int>dis;
queue<int>q;
int g[4][4];
inline bool inmap(int x,int y) {
    return x >=1 && x<= 3 && y >=1 && y<= 3;
}
int res = 0;
int a;
void bfs() {
    q.push(a);
    q.push(es);
    mp[a] = 1,mp[es] = 2;
    dis[a] = 0,dis[es] = 0;
    while (q.size()) {
        auto u  = q.front();
        q.pop();
        bool f= false;
        int x,y;
        if (mp[u]==1)
            f = true;
        int uu = u;
        for (int i = 3; i>=1; i--) {
            for (int j = 3; j>=1; j--) {
                g[i][j] =uu%10;
                uu /=10;
                if (g[i][j]==0)
                    x=  i,y= j;
            }
        }
        for (int i = 0; i< 4; i++) {
            int nx =  x + dx[i];
            int ny =  y + dy[i];
            if (inmap(nx,ny) ) {
                int t = 0;
                swap(g[nx][ny],g[x][y]);

                for (int a = 1; a <= 3; a++) {
                    for (int b = 1; b<=3 ;b++) {
                        t = t * 10 + g[a][b];
                    }
                }
                if (mp[t]+ mp[u] == 3) {
                    res = dis[t] + dis[u] + 1;
                    return ;
                }
                if (mp[t] == mp[u]) {
                    swap(g[nx][ny],g[x][y]);
                    continue;
                }
                if (f)
                    mp[t] =  1;
                else
                    mp[t] = 2;
                dis[t] = dis[u] + 1;
                q.push(t);
                swap(g[nx][ny],g[x][y]);
            }
        }
    }
}
int main(){
    cin>>a;
    bfs();
    cout<<res;
    return 0;
}

Astar

#include<bits/stdc++.h>
using namespace std;
int ntx[] = {1,0,0,0,1,2,2,2,1};
int nty[] = {1,0,1,2,2,2,1,0,0};
int f(string s) {
    int ans = 0;
    for (int i = 0; i<s.size(); i++) {
        int t = s[i]-'0';
        ans += abs(i/3-ntx[t]) + abs(i%3-nty[t]);
    }
    return ans;
}

typedef pair<int,string> PIS;
string s;
int dx[] ={-1,0,1,0};
int dy[] ={0,-1,0,1};
unordered_map<string,int>d;
int bfs() {
    priority_queue<PIS>q;
    q.push({-f(s),s});
    d[s] = 0;
    while (q.size()) {
        auto u = q.top();
        q.pop();
        int id = 0;
        if (u.second == "123804765")
            return d[u.second];
        for (int i = 0 ; i< u.second.size() ;i++) {
            if (u.second[i] =='0') {
                id = i;break;
            }
        }
        for (int i= 0; i<  4; i++) {
            int nx = id/3 + dx[i];
            int ny = id%3 + dy[i];
            if (nx>=0 && nx<3 && ny>=0 && ny<3) {
                string tmp = u.second;
                int nid = nx*3 + ny;
                swap(tmp[id],tmp[nid]);
                if (!d.count(tmp)) {
                    d[tmp] = d[u.second] +1;
                    q.push({-f(tmp) - d[tmp],tmp});
                }
            }
        }
    }
    return -1;
}
int main() {
    cin>>s;
    cout<<bfs();

    return 0;
}