hdu 5012 Dice 记忆化搜索
2014-09-14 20:21
666 查看
Dice
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 152 Accepted Submission(s): 87
Problem Description
There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a1.a2,a3,a4,a5,a6 to be numbers written on top face, bottom face, left face, right face,
front face and back face of dice A. Similarly, consider b1.b2,b3,b4,b5,b6 to be numbers on specific faces of dice B. It’s guaranteed that all numbers written on dices are integers no smaller
than 1 and no more than 6 while ai ≠ aj and bi ≠ bj for all i ≠ j. Specially, sum of numbers on opposite faces may not be 7.
At the beginning, the two dices may face different(which means there exist some i, ai ≠ bi). Ddy wants to make the two dices look the same from all directions(which means for all i, ai = bi) only by the following
four rotation operations.(Please read the picture for more information)
![](https://oscdn.geek-share.com/Uploads/Images/Content/201910/23/4d5d5c19f17f92722270306477e25374.jpg)
Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal.
Input
There are multiple test cases. Please process till EOF.
For each case, the first line consists of six integers a1,a2,a3,a4,a5,a6, representing the numbers on dice A.
The second line consists of six integers b1,b2,b3,b4,b5,b6, representing the numbers on dice B.
Output
For each test case, print a line with a number representing the answer. If there’s no way to make two dices exactly the same, output -1.
Sample Input
1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 5 6 4 3 1 2 3 4 5 6 1 4 2 5 3 6
Sample Output
0 3 -1
把骰子的初始状态记录下来。步数初始化为0;
这样不停得按他所说的四种转法记忆化+bfs就行了。 一般这种搜步数的我都用bfs,不知道dfs行不行。
中间在结构体中打的是括号的重载运算符操作。不知道为什么结构体中用数组不能用这个方法来构造。
size记录下步数,遇到重点状态,然后把size输出出来就好了。
#include<stdio.h> #include<queue> using namespace std; struct saizi { int b1,b2,b3,b4,b5,b6; int size; saizi() {} saizi(int a1,int a2,int a3,int a4,int a5,int a6,int si):b1(a1),b2(a2),b3(a3),b4(a4),b5(a5),b6(a6),size(si){}; /*saizi(int a1,int a2,int a3,int a4,int a5,int a6,int si) { b1=a1; a[2]=a2; a[3]=a3; a[4]=a4; a[5]=a5; a[6]=a6; size=si; }*/ bool operator == (const saizi& b) const // 如果 是普通的优先队列 运算符是< { if(b.b1==b1&&b.b2==b2&&b.b3==b3&&b.b4==b4&&b.b5==b5&&b.b6==b6) return 1; return 0; } }sai,tem,tel; int dice[10][10][10][10][10][10]; int a[10]; int bfs() { //if(dice[a [1]][a [2]][a [3]][a [4]][a [5]][a [6]]) //return ; queue<saizi> q; sai.size=0; q.push(sai); while(!q.empty()) { sai=q.front(); q.pop(); if(sai==tel) return sai.size; if(dice[sai.b1][sai.b2][sai.b3][sai.b4][sai.b5][sai.b6]==1) continue; dice[sai.b1][sai.b2][sai.b3][sai.b4][sai.b5][sai.b6]=1; //左为轴 //shun1 tem=saizi(sai.b6,sai.b5,sai.b3,sai.b4,sai.b1,sai.b2,sai.size+1); q.push(tem); //shun2 //tem=saizi(sai.b2,sai.b1,sai.b3,sai.b4,sai.b6,sai.b5,sai.size+1); //q.push(tem); //shun3 tem=saizi(sai.b5,sai.b6,sai.b3,sai.b4,sai.b2,sai.b1,sai.size+1); q.push(tem); //前为轴 //shun1 tem=saizi(sai.b3,sai.b4,sai.b2,sai.b1,sai.b5,sai.b6,sai.size+1); q.push(tem); //shun2 //tem=saizi(sai.b2,sai.b1,sai.b4,sai.b3,sai.b5,sai.b6,sai.size+1); //q.push(tem); //shun3 tem=saizi(sai.b4,sai.b3,sai.b1,sai.b2,sai.b5,sai.b6,sai.size+1); q.push(tem); //上为轴 //顺1 //tem=saizi(sai.b1,sai.b2,sai.b5,sai.b6,sai.b4,sai.b3,sai.size+1); //q.push(tem); /* //shun2 tem=saizi(sai.b1,sai.b2,sai.b3,sai.b4,sai.b6,sai.b5,sai.size+1); q.push(tem); //shun3 tem=saizi(sai.b1,sai.b2,sai.b6,sai.b5,sai.b3,sai.b4,sai.size+1); q.push(tem); */ } return -1; } int main() { int ans; while(scanf("%d",&sai.b1)!=EOF) { memset(dice,0,sizeof(dice)); scanf("%d",&sai.b2); scanf("%d",&sai.b3); scanf("%d",&sai.b4); scanf("%d",&sai.b5); scanf("%d",&sai.b6); scanf("%d",&tel.b1); scanf("%d",&tel.b2); scanf("%d",&tel.b3); scanf("%d",&tel.b4); scanf("%d",&tel.b5); scanf("%d",&tel.b6); ans=bfs(); printf("%d\n",ans); } return 0; }
相关文章推荐
- 【搜索】 HDU 5012 Dice
- hdu 5012 Dice(隐式图搜索)
- Dice - HDU 5012 搜索
- 【搜索】 HDU 5012 Dice
- HDU 5012 Dice 隐式图的搜索
- HDU 5012 Dice(西安网络赛F题)
- HDU 5012 Dice
- HDU 1016 Prime Ring Problem DFS + 记忆化搜索
- 【搜索】 HDOJ 5012 Dice
- HDU 5012 Dice (bfs + 记忆化搜索)
- hdu 5012——Dice
- hdu 5012 Dice BFS 2014 ACM/ICPC Asia Regional Xi'an Online
- HDU--杭电--1978--How many ways--记忆化搜索--这个鸟东西应该能算半个DP吧
- hdu 1142 记忆化搜索
- hdu 1978(记忆化DFS搜索)
- hdu 5012 Dice
- hdu(1331)(通过记忆化搜索减少程序运行时间)
- hdu 1142 记忆化搜索
- hdu1579 Function Run Fun 记忆化搜索启蒙题
- HDU5012 Dice