洛谷 P1379 八数码难题(map && 双向bfs)

题目传送门

解题思路:

一道bfs,本题最难的一点就是如何储存已经被访问过的状态,如果直接开一个bool数组,空间肯定会炸,所以我们要用另一个数据结构存,STL大法好,用map来存,直接AC.

AC代码:

 #include<cstdio>
 #include<iostream>
 #include<map>
 #include<queue> 
 
 using namespace std;
 
 int a[][],n;
 int ans = ;
 const int dx[]={-,,,},dy[]={,-,,};
 map<int,int> m;
 
 inline void bfs() {
     queue<long long> q;
     q.push(n);
     m[n] = ;
     while(!q.empty()) {
         int u = q.front();
         q.pop();
         int x = ,y= ,n = u;
         if(u == ans) break;
         for(int i = ;i >= ; i--)
             for(int j = ;j >= ; j--) {
                 a[i][j] = n % ;
                 n /= ;
                 if(!a[i][j]) x = i,y = j;
             }
         for(int i = ;i < ; i++) {
             int nx = x + dx[i],ny = y + dy[i],ns = ;
             if(nx <  || ny <  || nx >  || ny > ) continue;
             swap(a[nx][ny],a[x][y]);
             for(int i = ;i <= ; i++)
                 for(int j = ;j <= ; j++)
                     ns = ns *  + a[i][j];
             if(!m.count(ns)) {
                 m[ns] = m[u] + ;
                 q.push(ns);
             }
             swap(a[nx][ny],a[x][y]);
         }
     }
 }
 
 int main()
 {
     scanf("%d",&n);
     bfs();
     printf("%d",m[]);
     return ;
 }

普通bfs

emmm,虽然普通bfs能A,但还是感觉太慢,所以试了一下双向bfs.

 #include<cstdio>
 #include<iostream>
 #include<queue>
 #include<map>
 
 using namespace std;
 
 int a[][],n,n1 = ;
 int x,y;
 int wayx[] = {,-,,},wayy[] = {,,,-};
 map<int,int> p,sum;
 
 inline void juzhen(int o) {
     for(int i = ;i >= ; i--)
         for(int j = ;j >= ; j--) {
             a[i][j] = o % ;
             o /= ;
             if(a[i][j] == ) x = i,y = j;
         }
 }
 
 inline void double_bfs() {
     if(n1 == ) return ;
     queue<int> step[];
     step[].push(n);
     step[].push(n1);
     sum[n] = ;
     sum[n1] = ;
     p[n] = ;p[n1] = ;
     int deep = ,s;
     while(!step[].empty() || !step[].empty()) {
         int i = ;
         while(i < ) {
             if(sum[step[i].front()] != deep) {
                 i++;
                 continue;
             }
             s = step[i].front();
             step[i].pop();
             juzhen(s);
             for(int w = ;w < ; w++) {
                 int xx = x + wayx[w];
                 int yy = y + wayy[w];
                 if(xx <  || xx >  || yy <  || yy > ) continue;
                 swap(a[x][y],a[xx][yy]);
                 int pp = ;
                 for(int j = ;j <= ; j++)
                     for(int u = ;u <= ; u++)
                         pp = pp *  + a[j][u];
                 if(p.count(pp)) {
                     if(p[pp] ==  - i) {
                         if(p[pp] == )
                             printf("%d",(deep + ) * );
                         else
                             printf("%d",deep *  + );
                         return ;
                     }
                 }
                 else {
                     step[i].push(pp);
                     sum[pp] = sum[s] + ;
                     p[pp] = i;
                 }
                 swap(a[x][y],a[xx][yy]);
             }
         }
         deep++;
     }
 }
 
 inline void tepan() {
     if(n == n1) printf("");
     if(n == n1) n = n1 = ;
 }
 
 int main()
 {
     scanf("%d",&n);
     tepan();
     double_bfs();
     return ;
 }

双向bfs

双向bfs确实快很多很多,普通bfs耗时7.38s,而双向bfs只耗时291ms,快了25倍!!!!!!