Dice
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 491 Accepted Submission(s): 290
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 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 5 6 4 3
1 2 3 4 5 6
1 4 2 5 3 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 5 6 4 3
1 2 3 4 5 6
1 4 2 5 3 6
Sample Output
0
3
-1
3
-1
Source
Recommend
卜神的代码
为了测试康托展开， 文中注释部分是卜神原来的代码，也是ac的。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std; typedef long long ll; const int CANTO = ;
const int LEN = ; int fac[];
bool vis[CANTO];
int begin, end;
struct Sit
{
int arr[LEN];
int step;
}; void makefac()
{
fac[] = fac[] = ;
for(int i = ; i <= ; i++)
fac[i] = i * fac[i-];
} int canto(int arr[])
{
int res = ;
for(int i = ; i < LEN; i++){
int num=;
for(int j=i+;j<LEN;j++)
if(arr[j]<arr[i]) num++;
res+=(num*fac[LEN-i-]);
}
// res += fac[i+2] * arr[i];
return res;
} int bfs(Sit src)
{
queue <Sit> q;
Sit now, tmp;
int t, c;
vis[begin] = true;
q.push(src);
while(!q.empty())
{
now = q.front();
q.pop();
tmp = now, tmp.step++; t = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = t;
c = canto(tmp.arr);
if (c == end){
return tmp.step;
}
if (!vis[c]){
vis[c] = true;
q.push(tmp);
}
tmp = now, tmp.step++; t = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = t;
c = canto(tmp.arr);
if (c == end){
return tmp.step;
}
if (!vis[c]){
vis[c] = true;
q.push(tmp);
} tmp = now, tmp.step++; t = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = t;
c = canto(tmp.arr);
if (c == end){
return tmp.step;
}
if (!vis[c]){
vis[c] = true;
q.push(tmp);
} tmp = now, tmp.step++; t = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = tmp.arr[];
tmp.arr[] = t;
c = canto(tmp.arr);
if (c == end){
return tmp.step;
}
if (!vis[c]){
vis[c] = true;
q.push(tmp);
}
}
return -;
} int main()
{
makefac();
int src[LEN];
int dst[LEN];
while(~scanf("%d", src)){
memset(vis, , sizeof(vis));
for(int i = ; i < LEN; i++)
scanf("%d", src+i);
for(int i = ; i < LEN; i++)
scanf("%d", dst+i);
begin = canto(src);
end = canto(dst);
if (begin == end){
printf("0\n");
continue;
}
Sit x;
memcpy(x.arr, src, LEN*sizeof(int));
x.step = ;
printf("%d\n", bfs(x));
}
return ;
}
经过以下代码验证，网上的代码，康托展开是连续值，卜神的只是起到了展开作用，不过代码更加简洁。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std; typedef long long ll; const int CANTO = ;
const int LEN = ; int fac[];
int a[LEN]; void makefac()
{
fac[] = fac[] = ;
for(int i = ; i <= ; i++)
fac[i] = i * fac[i-];
} int canto(int arr[])
{
int res = ;
for(int i = ; i < LEN; i++){
int num=;
for(int j=i+;j<LEN;j++)
if(arr[j]<arr[i]) num++;
res+=(num*fac[LEN-i-]);
}
// res += fac[i+2] * arr[i];
return res;
} int canto2(int arr[])
{
int res = ;
for(int i = ; i < LEN; i++){
// int num=0;
// for(int j=i+1;j<LEN;j++)
// if(arr[j]<arr[i]) num++;
// res+=(num*fac[LEN-i-1]);
res += fac[i+] * arr[i];
}
// res += fac[i+2] * arr[i];
return res;
} int main()
{
freopen("data.out","w",stdout);
makefac();
a[]=;a[]=;a[]=;a[]=;a[]=;a[]=;
printf("a=%d %d %d %d %d %d ",a[],a[],a[],a[],a[],a[]);
printf("can=%d %d\n",canto(a),canto2(a));
while(next_permutation(a+, a + LEN) ){
printf("a=%d %d %d %d %d %d ",a[],a[],a[],a[],a[],a[]);
printf("can=%d %d\n",canto(a),canto2(a));
}
return ;
}
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