HDU - 5012 Dice(BFS)
来源:互联网 发布:希捷 数据恢复服务 编辑:程序博客网 时间:2024/05/16 02:04
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)
Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal.
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)
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.
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 61 2 3 4 5 61 2 3 4 5 61 2 5 6 4 31 2 3 4 5 61 4 2 5 3 6
Sample Output
03-1题意:求起始到终态的步骤思路:简单的BFS+判重,注意判重数组开准点#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int maxn = 6;struct Node{int arr[maxn], step;Node() {memset(arr, 0, sizeof(arr));step = 0; }}start, end;int vis[maxn*200000];int cal(Node a) {int num = 0;for (int i = 0; i < maxn; i++) {num = num * 10 + a.arr[i]; }return num;}bool equal(Node a, Node b) {for (int i = 0; i < maxn; i++)if (a.arr[i] != b.arr[i])return false;return true;}Node turn(Node a, int i) {Node c;if (i == 1) {c.arr[0] = a.arr[3];c.arr[1] = a.arr[2];c.arr[2] = a.arr[0];c.arr[3] = a.arr[1];c.arr[4] = a.arr[4];c.arr[5] = a.arr[5]; }if (i == 2) {c.arr[0] = a.arr[2];c.arr[1] = a.arr[3];c.arr[2] = a.arr[1];c.arr[3] = a.arr[0]; c.arr[4] = a.arr[4];c.arr[5] = a.arr[5]; }if (i == 3) {c.arr[0] = a.arr[5];c.arr[1] = a.arr[4];c.arr[2] = a.arr[2];c.arr[3] = a.arr[3]; c.arr[4] = a.arr[0];c.arr[5] = a.arr[1]; }if (i == 4) {c.arr[0] = a.arr[4];c.arr[1] = a.arr[5];c.arr[2] = a.arr[2];c.arr[3] = a.arr[3]; c.arr[4] = a.arr[1];c.arr[5] = a.arr[0];}return c;}int bfs() {memset(vis, 0, sizeof(vis));queue<Node> q;q.push(start);Node tmp; vis[cal(start)] = 1;while (!q.empty()) {tmp = q.front(); q.pop(); if (equal(tmp, end)) {return tmp.step;}for (int i = 1; i <= 4; i++) {Node c;c = turn(tmp, i);if (!vis[cal(c)]) {c.step = tmp.step + 1; vis[cal(c)] = 1;q.push(c);} }} return -1;}int main() {while (scanf("%d", &start.arr[0]) != EOF) {for (int i = 1; i < maxn; i++) scanf("%d", &start.arr[i]);for (int i = 0; i < maxn; i++) scanf("%d", &end.arr[i]);printf("%d\n", bfs());} return 0;}
1 0
- HDU 5012 Dice (bfs)
- HDU - 5012 Dice(BFS)
- HDU-#5012 Dice(BFS)
- hdu 5012 Dice(bfs)
- HDU 5012 Dice (BFS)
- HDU 5012-Dice(BFS)
- HDU 5012 Dice bfs
- HDU - 5012 Dice BFS
- bfs-HDU 5012Dice
- HDU - 5012 Dice(bfs+hash)
- HDU 5012 Dice 普通bfs
- HDU 5012 Dice (bfs + 记忆化搜索)
- BFS + 剪枝 之 hdu 5012 Dice
- hdu HDU5012-Dice(BFS)
- hdu 5012 Dice(西安网络赛F题,BFS)
- hdu 5012 Dice
- HDU 5012 Dice
- hdu 5012 dice
- 有两个指针pa,pb分别指向有两个数,a,b,请写一个函数交换两个指针的指向,也就是让pa指向b,让pb指向a
- HDOJ 5007 Post Robot
- 获取路由器超级用户权限(你懂得...)
- Maximum Subarray
- 实现一个计算器
- HDU - 5012 Dice(BFS)
- python的优先权队列
- 图片像素对比OpenCV实现,实现人工分割跟算法分割图像结果的对比
- 图的深度遍历(DFS)
- 纪念逝去的昨天(1)
- Java学习笔记之Java基础
- html中input文本框,初始里边有文字提示,当点击时,文字消失
- 公司笔试题——统计10进制数中的二进制值有多少个1
- Vistual studio 快捷键积累