|-------| |-------| |-------| | | | | | | | |---O | |---O | | O | | | | | | | |-------| |-------| |-------| A B C |-------| |-------| |-------| | | | | | | | O | | O | | O | | | | | | | | | | |-------| |-------| |-------| D E F |-------| |-------| |-------| | | | | | | | O | | O---| | O | | | | | | | | | |-------| |-------| |-------| G H I
The goal is to find a minimal sequence of moves to return all the dials to 12 o'clock. Nine different ways to turn the dials on the clocks are supplied via a table below; each way is called a move. Select for each move a number 1 through 9 which will cause the dials of the affected clocks (see next table) to be turned 90 degrees clockwise.
Move | Affected clocks |
1 | ABDE |
2 | ABC |
3 | BCEF |
4 | ADG |
5 | BDEFH |
6 | CFI |
7 | DEGH |
8 | GHI |
9 | EFHI |
Example
Each number represents a time accoring to following table:9 9 12 9 12 12 9 12 12 12 12 12 12 12 12 6 6 6 5 -> 9 9 9 8-> 9 9 9 4 -> 12 9 9 9-> 12 12 12 6 3 6 6 6 6 9 9 9 12 9 9 12 12 12
[But this might or might not be the `correct' answer; see below.]
PROGRAM NAME: clocks
INPUT FORMAT
Lines 1-3: | Three lines of three space-separated numbers; each number represents the start time of one clock, 3, 6, 9, or 12. The ordering of the numbers corresponds to the first example above. |
SAMPLE INPUT (file clocks.in)
9 9 12 6 6 6 6 3 6
OUTPUT FORMAT
A single line that contains a space separated list of the shortest sequence of moves (designated by numbers) which returns all the clocks to 12:00. If there is more than one solution, print the one which gives the lowest number when the moves are concatenated (e.g., 5 2 4 6 < 9 3 1 1).
SAMPLE OUTPUT (file clocks.out)
4 5 8 9
此题大概提交了十几次。备份了三个。再加上拷别人的标答。刷新了做usaco题的悲剧记录。
到目前为止对题解的神展开理解不能。
就……就这样吧……
悲剧的只过了两组数据的bfs:
/*
ID: wtff0411
PROG: clocks
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <cstring>
#include <vector>
#include <cmath>
#include <queue>
using namespace std;
char ans[100];
long long move[10]={0,18911232,19136512, 2363904, 16810048, 2134536, 262657, 36936, 73, 4617};
struct p{
friend bool operator <(p x,p y)
{
return strcmp(x.res,y.res)>0;
}
long long num;
char res[100];
};
priority_queue <p> q;
void solve(int n)
{
p pp;
pp.num=n;
int i;
int j=10000;
pp.res[0]='0';
q.push(pp);
while(1)
{
//cout<<q.top().num<<endl;
if(q.top().num==0)
system("pause");
p t=q.top();
q.pop();
if(t.num==0)
{
strcpy(ans,t.res);
return ;
}
for(i=1;i<10;i++)
{
p tt;
tt.num=(t.num+move[i])&57521883;
strcpy(tt.res,t.res);
tt.res[0]++;
tt.res[tt.res[0]-'0']=i+'0';
q.push(tt);
}
}
}
int main()
{
//freopen("clocks.in","r",stdin);
//freopen("clocks.out","w",stdout);
int i,j;
int input;
memset(ans,'0',sizeof(ans));
int origin=0;
for(i=8;i>=0;i--)
{
cin>>input;
origin+=((input/3)%4)*(int)pow(8.0,i);
}
solve(origin);
/*int num;
for(i=1;;i*=10)
if(res/i==0)
{i=i/10;break;}
for(;i>=10;i/=10)
{
cout<<res/i<<" ";
res%=i;
}
cout<<res;*/
// cout<<ans;
for(i=1;i<ans[0]-'0';i++)
{
cout<<ans[i]<<" ";
}
cout<<ans[i];
cout<<endl;
system("pause");
return 0;
}
干!还特地去研究了优先队列了呢(虽然以前也做到过,但这次第一次是排自己写的结构啊)。难道是东西太多队列满了- -?烦烦烦烦烦。