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
;
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]
= {{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;
}