HDU 4801 Pocket Cube bfs

Pocket Cube

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 816    Accepted Submission(s): 255

Problem Description

Pocket Cube is a 3-D combination puzzle. It is a 2 × 2 × 2 cube, which means it is constructed by 8 mini-cubes. For a combination of 2 × 2 mini-cubes which sharing a whole cube face, you can twist it 90 degrees in clockwise or counterclockwise direction, this twist operation is called one twist step.
Considering all faces of mini-cubes, there will be totally 24 faces painted in 6 different colors (Indexed from 0), and there will be exactly 4 faces painted in each kind of color. If 4 mini-cubes' faces of same color rely on same large cube face, we can call the large cube face as a completed face.



Now giving you an color arrangement of all 24 faces from a scrambled Pocket Cube, please tell us the maximum possible number of completed faces in no more than N twist steps.
Index of each face is shown as below:

 

Input

There will be several test cases. In each test case, there will be 2 lines. One integer N (1 ≤ N ≤ 7) in the first line, then 24 integers Ci separated by a single space in the second line. For index 0 ≤ i < 24, Ci is color of the corresponding face. We guarantee that the color arrangement is a valid state which can be achieved by doing a finite number of twist steps from an initial cube whose all 6 large cube faces are completed faces.

 

Output

For each test case, please output the maximum number of completed faces during no more than N twist step(s).

 

Sample Input

10 0 0 0 1 1 2 2 3 3 1 1 2 2 3 3 4 4 4 4 5 5 5 510 4 0 4 1 1 2 5 3 3 1 1 2 5 3 3 4 0 4 0 5 2 5 2

 

Sample Output

62

 

Source

2013 Asia Changsha Regional Contest

题意：给出初始和目标魔方状态，问最短路。

思路：其实不难。 不过你要注意到只需要旋转三个面就可以了。

代码：

#include <iostream>#include <vector>#include <algorithm>#include <string.h>#include <cstring>#include <time.h>#include <stdio.h>#include <queue>#include <cmath>#include <set>#include <math.h>#define rep(i,a,b) for(int i=(a);i<(b);++i)#define rrep(i,b,a) for(int i = (b); i >= (a); --i)#define clr(a,x) memset(a,(x),sizeof(a))#define LL long long#define eps 1e-10#define zero(x) -eps < (x) && (x) < epsusing namespace std;int face[6][4] ={    {0,1,3,2},    {4,5,11,10},    {6,7,13,12},    {8,9,15,14},    {16,17,19,18},    {20,21,23,22}};int e[6][8] ={    {9,8,7,6,5,4,22,23},    {0,2,6,12,16,18,20,22},    {2,3,8,14,17,16,11,5},    {23,21,19,17,13,7,3,1},    {10,11,12,13,14,15,21,20},    {18,19,15,9,1,0,4,10}};struct State{    char a[24];    bool operator == (const State & st) const    {        rep(i,0,24)        if (a[i] != st.a[i]) return false;        return true;    }    bool operator < (const State & st) const    {        rep(i,0,24)        if (a[i] != st.a[i]) return a[i] < st.a[i];        return false;    }    void out() const    {        printf("\n    ");        printf("%2c%2c\n",a[0],a[1]);        printf("    ");        printf("%2c%2c\n",a[2],a[3]);        rep(i,4,10) printf("%2c",a[i]);        puts("");        rep(i,10,16) printf("%2c",a[i]);        puts("");        printf("    ");        printf("%2c%2c\n",a[16],a[17]);        printf("    ");        printf("%2c%2c\n",a[18],a[19]);        printf("    ");        printf("%2c%2c\n",a[20],a[21]);        printf("    ");        printf("%2c%2c\n\n",a[22],a[23]);    }    int d;    int completeFace() const    {        int cnt = 0;        if (a[0] == a[1] && a[1] == a[2]            && a[2] == a[3]) ++cnt;        if (a[6] == a[7] && a[7] == a[12]            && a[12] == a[13]) ++cnt;        if (a[8] == a[9] && a[9] == a[14]            && a[14] == a[15]) ++cnt;        if (a[4] == a[5] && a[5] == a[10]            && a[10] == a[11]) ++cnt;        if (a[16] == a[17] && a[17] == a[18]            && a[18] == a[19]) ++cnt;        if (a[20] == a[21] && a[21] == a[22]            && a[22] == a[23]) ++cnt;        return cnt;    }    void Rotate(vector<char*>& face,vector<char*>& edge)    {        char ch = *face.back();        rrep(i,face.size()-1,1) {            *face[i] = *face[i-1];        }        *face[0] = ch;        char ch1 = *edge[6];        char ch2 = *edge[7];        rrep(i,edge.size()-1,2) {            *edge[i] = *edge[i-2];        }        *edge[0] = ch1;        *edge[1] = ch2;    }}start;int n;void input(){    rep(i,0,24) {        int c; scanf("%d",&c);        start.a[i] = c + '0';    }}inline int max(int a,int b) { return a > b ? a : b; }void solve(){    int ans = 0;    queue<State> q;    set<State> vis;    vis.insert(start);    start.d = 0;    q.push(start);    vector<char*> a,b;    State now,t;    while (q.size()) {        now = q.front(); q.pop();    //    printf("%d %d\n",q.size(),now.d);        int h = now.completeFace();        if (h > ans) ans = h;      //  if (h + (n - now.d) * 2 <= ans) continue;        if (ans == 6) break;        if (now.d >= n) continue;        int cnt = 0;        rep(i,0,3) {            a.clear(); b.clear();            t = now;            rep(j,0,4)                a.push_back(&t.a[face[i][j]]);            rep(j,0,8)                b.push_back(&t.a[e[i][j]]);         //   now.out();            ++t.d;            t.Rotate(a,b);         //   t.out();            if (!vis.count(t)) {                q.push(t);                vis.insert(t);                ++cnt;            }            a.clear(); b.clear();            t = now; ++t.d;            rrep(j,3,0)                a.push_back(&t.a[face[i][j]]);            rrep(j,7,0)                b.push_back(&t.a[e[i][j]]);            t.Rotate(a,b);         //   now.out();          //  t.out();            if (!vis.count(t)) {                q.push(t);                vis.insert(t);                ++cnt;            }        }        //cout << cnt << endl;    }    printf("%d\n",ans);}int main(){    #ifdef ACM        freopen("in.txt", "r", stdin);    #endif // AC    while (scanf("%d",&n)==1) {        input();        solve();    }}