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:
1 #include<cstdio> 2 #include<iostream> 3 #include<map> 4 #include<queue> 5 6 using namespace std; 7 8 int a[3][3],n; 9 int ans = 123804765; 10 const int dx[]={-1,0,0,1},dy[]={0,-1,1,0}; 11 map<int,int> m; 12 13 inline void bfs() { 14 queue<long long> q; 15 q.push(n); 16 m[n] = 0; 17 while(!q.empty()) { 18 int u = q.front(); 19 q.pop(); 20 int x = 0,y= 0,n = u; 21 if(u == ans) break; 22 for(int i = 2;i >= 0; i--) 23 for(int j = 2;j >= 0; j--) { 24 a[i][j] = n % 10; 25 n /= 10; 26 if(!a[i][j]) x = i,y = j; 27 } 28 for(int i = 0;i < 4; i++) { 29 int nx = x + dx[i],ny = y + dy[i],ns = 0; 30 if(nx < 0 || ny < 0 || nx > 2 || ny > 2) continue; 31 swap(a[nx][ny],a[x][y]); 32 for(int i = 0;i <= 2; i++) 33 for(int j = 0;j <= 2; j++) 34 ns = ns * 10 + a[i][j]; 35 if(!m.count(ns)) { 36 m[ns] = m[u] + 1; 37 q.push(ns); 38 } 39 swap(a[nx][ny],a[x][y]); 40 } 41 } 42 } 43 44 int main() 45 { 46 scanf("%d",&n); 47 bfs(); 48 printf("%d",m[123804765]); 49 return 0; 50 }
emmm,虽然普通bfs能A,但还是感觉太慢,所以试了一下双向bfs.
1 #include<cstdio> 2 #include<iostream> 3 #include<queue> 4 #include<map> 5 6 using namespace std; 7 8 int a[4][4],n,n1 = 123804765; 9 int x,y; 10 int wayx[4] = {1,-1,0,0},wayy[4] = {0,0,1,-1}; 11 map<int,int> p,sum; 12 13 inline void juzhen(int o) { 14 for(int i = 3;i >= 1; i--) 15 for(int j = 3;j >= 1; j--) { 16 a[i][j] = o % 10; 17 o /= 10; 18 if(a[i][j] == 0) x = i,y = j; 19 } 20 } 21 22 inline void double_bfs() { 23 if(n1 == 0) return ; 24 queue<int> step[2]; 25 step[0].push(n); 26 step[1].push(n1); 27 sum[n] = 0; 28 sum[n1] = 0; 29 p[n] = 0;p[n1] = 1; 30 int deep = 0,s; 31 while(!step[0].empty() || !step[1].empty()) { 32 int i = 0; 33 while(i < 2) { 34 if(sum[step[i].front()] != deep) { 35 i++; 36 continue; 37 } 38 s = step[i].front(); 39 step[i].pop(); 40 juzhen(s); 41 for(int w = 0;w < 4; w++) { 42 int xx = x + wayx[w]; 43 int yy = y + wayy[w]; 44 if(xx < 1 || xx > 3 || yy < 1 || yy > 3) continue; 45 swap(a[x][y],a[xx][yy]); 46 int pp = 0; 47 for(int j = 1;j <= 3; j++) 48 for(int u = 1;u <= 3; u++) 49 pp = pp * 10 + a[j][u]; 50 if(p.count(pp)) { 51 if(p[pp] == 1 - i) { 52 if(p[pp] == 0) 53 printf("%d",(deep + 1) * 2); 54 else 55 printf("%d",deep * 2 + 1); 56 return ; 57 } 58 } 59 else { 60 step[i].push(pp); 61 sum[pp] = sum[s] + 1; 62 p[pp] = i; 63 } 64 swap(a[x][y],a[xx][yy]); 65 } 66 } 67 deep++; 68 } 69 } 70 71 inline void tepan() { 72 if(n == n1) printf("0"); 73 if(n == n1) n = n1 = 0; 74 } 75 76 int main() 77 { 78 scanf("%d",&n); 79 tepan(); 80 double_bfs(); 81 return 0; 82 }
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!!!!!!