hdu 5012(bfs)

题意：给出两个正方体，每个正方体都有6种颜色，问是否可以经过左转，右转，前转，后转四种转法让两个正方体看起来一样(对应面颜色相同)，最少转几次。

题解：bfs水题。

#include <stdio.h>
#include <queue>
#include <map>
using namespace std;
const int N = 7;
struct Cube {
    int c[N];
    int step;
    bool operator == (Cube a) const {
        for (int i = 0; i < 6; i++) {
            if (this -> c[i] != a.c[i])
                return false;
        }
        return true;
    }
}A, B;
int flag[4][N] = {{3, 2, 0, 1, 4, 5}, {2, 3, 1, 0, 4, 5}, {5, 4, 2, 3, 0, 1}, {4, 5, 2, 3, 1, 0}};
queue<Cube> q;
map<int, int> m;

bool Hash(Cube u) {
    int temp = 0;
    for (int i = 0; i < 6; i++)
        temp = temp * 10 + u.c[i];
    if (!m[temp]) {
        m[temp] = 1;
        return true;
    }
    return false;
}

void bfs() {
    while (!q.empty())
        q.pop();
    A.step = 0;
    q.push(A);
    Hash(q.front());
    while (!q.empty()) {
        Cube u = q.front();
        q.pop();
        Cube temp;
        for (int j = 0; j < 4; j++) {
            for (int i = 0; i < 6; i++)
                temp.c[i] = u.c[flag[j][i]];    
            temp.step = u.step + 1;
            if (Hash(temp)) {
                if (temp == B) {
                    printf("%d\n", temp.step);    
                    return;
                }
                q.push(temp);
            }
        }
    }
    printf("-1\n");
}

int main() {
    while (scanf("%d", &A.c[0]) == 1) {
        m.clear();
        for (int i = 1; i < 6; i++)
            scanf("%d", &A.c[i]);
        for (int i = 0; i < 6; i++)
            scanf("%d", &B.c[i]);
        if (A == B) {
            printf("0\n");
            continue;
        }
        bfs();
    }
    return 0;
}