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  • BFS + 剪枝 之 hdu 5012 Dice

    /*
    BFS 遍历所有状态,当寻找到要转到的状态,输出BFS的层数,即答案,否则输出 -1;
    注意:
    剪枝: 当BFS到某一种状态时,若此状态已在前面所寻找的状态中出现过,则不用再继续从此状态BFS了,因为此状态所能寻找的状态已经在之前都处理过了。
    */
      1 #include <iostream>
      2 #include <cstdlib>
      3 #include <cstdio>
      4 #include <queue>
      5 #include <vector>
      6 #include <map>
      7 using namespace std;
      8 struct myNode {
      9     int Top, Bottom, Left, Right, Front, Back;
     10 };
     11 typedef pair<myNode, int> P;
     12 
     13 myNode node[2];
     14 vector<myNode> vec;
     15 
     16 void myDeal(int myCase, myNode myNd, myNode &t_Nd) {
     17     // top face, bottom face, left face, right face, front face and back face
     18     switch (myCase) {
     19     case 1: {
     20         t_Nd.Top = myNd.Right;
     21         t_Nd.Bottom = myNd.Left;
     22         t_Nd.Left = myNd.Top;
     23         t_Nd.Right = myNd.Bottom;
     24         t_Nd.Front = myNd.Front;
     25         t_Nd.Back = myNd.Back;
     26         return;
     27     }
     28     case 2: {
     29         t_Nd.Top = myNd.Left;
     30         t_Nd.Bottom = myNd.Right;
     31         t_Nd.Left = myNd.Bottom;
     32         t_Nd.Right = myNd.Top;
     33         t_Nd.Front = myNd.Front;
     34         t_Nd.Back = myNd.Back;
     35         return;
     36     }
     37     case 3: {
     38         t_Nd.Top = myNd.Back;
     39         t_Nd.Bottom = myNd.Front;
     40         t_Nd.Front = myNd.Top;
     41         t_Nd.Back = myNd.Bottom;
     42         t_Nd.Left = myNd.Left;
     43         t_Nd.Right = myNd.Right;
     44         return;
     45     }
     46     case 4: {
     47         t_Nd.Top = myNd.Front;
     48         t_Nd.Bottom = myNd.Back;
     49         t_Nd.Front = myNd.Bottom;
     50         t_Nd.Back = myNd.Top;
     51         t_Nd.Left = myNd.Left;
     52         t_Nd.Right = myNd.Right;
     53         return;
     54     }
     55     default:
     56         system("pause");
     57         return;
     58     }
     59 }
     60 
     61 bool myJudge(const myNode n1, const myNode n2) {
     62     if ((n1.Top == n2.Top) && (n1.Bottom == n2.Bottom) && (n1.Front == n2.Front) &&
     63         (n1.Back == n2.Back) && (n1.Left == n2.Left) && (n1.Right == n2.Right)) {
     64         return true;
     65     }
     66     return false;
     67 }
     68 
     69 bool Judge(myNode nd) {
     70     for (int i = 0; i < vec.size(); ++i) {
     71         if (myJudge(vec[i], nd)) {
     72             return true;
     73         }
     74     }
     75     return false;
     76 }
     77 
     78 void Solve() {
     79     queue<P> que;
     80     que.push(P(node[0], 0));
     81     while (!que.empty()) {
     82         P p1 = que.front();
     83         if (Judge(p1.first)){
     84             //cout << "Yes" << endl;
     85             que.pop();
     86             continue;
     87         }
     88         vec.push_back(p1.first);
     89         que.pop();
     90         if (myJudge(p1.first, node[1])) {
     91             printf("%d
    ", p1.second);
     92             return;
     93         }
     94         myNode t_Nd;
     95         for (int i = 1; i <= 4; ++i) {
     96             myDeal(i, p1.first, t_Nd);
     97             //cout << t_Nd.Top << " " << t_Nd.Bottom << " " << t_Nd.Left << " " << t_Nd.Right << " " << t_Nd.Front << " " << t_Nd.Back << endl;
     98             que.push(P(t_Nd, p1.second + 1));
     99         }
    100     }
    101     printf("-1
    ");
    102 }
    103 
    104 int main() {
    105     //freopen("input.txt", "r", stdin);
    106     int n;
    107     while (~scanf("%d %d %d %d %d %d", &node[0].Top, &node[0].Bottom, &node[0].Left, &node[0].Right, &node[0].Front, &node[0].Back)) {
    108         scanf("%d %d %d %d %d %d", &node[1].Top, &node[1].Bottom, &node[1].Left, &node[1].Right, &node[1].Front, &node[1].Back);
    109         Solve();
    110         if (!vec.empty()) vec.clear();
    111     }
    112     return 0;
    113 }
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  • 原文地址:https://www.cnblogs.com/shijianming/p/4140802.html
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