USACO 2.2.4 Party Lamps 解题报告
Party Lamps
IOI 98
To brighten up the gala dinner of the IOI'98 we have a set of N (10 <= N <= 100) colored lamps numbered from 1 to N.
The lamps are connected to four buttons:
- Button 1: When this button is pressed, all the lamps change their state: those that are ON are turned OFF and those that are OFF are turned ON.
- Button 2: Changes the state of all the odd numbered lamps.
- Button 3: Changes the state of all the even numbered lamps.
- Button 4: Changes the state of the lamps whose number is of the form 3xK+1 (with K>=0), i.e., 1,4,7,...
A counter C records the total number of button presses.
When the party starts, all the lamps are ON and the counter C is set to zero.
You are given the value of counter C (0 <= C <= 10000) and the final state of some of the lamps after some operations have been executed. Write a program to determine all the possible final configurations of the N lamps that are consistent with the given information, without repetitions.
PROGRAM NAME: lamps
INPUT FORMAT
No lamp will be listed twice in the input.
Line 1:
N
Line 2:
Final value of C
Line 3:
Some lamp numbers ON in the final configuration, separated by one space and terminated by the integer -1.
Line 4:
Some lamp numbers OFF in the final configuration, separated by one space and terminated by the integer -1.
SAMPLE INPUT (file lamps.in)
10 1 -1 7 -1
In this case, there are 10 lamps and only one button has been pressed. Lamp 7 is OFF in the final configuration.
OUTPUT FORMAT
Lines with all the possible final configurations (without repetitions) of all the lamps. Each line has N characters, where the first character represents the state of lamp 1 and the last character represents the state of lamp N. A 0 (zero) stands for a lamp that is OFF, and a 1 (one) stands for a lamp that is ON. The lines must be ordered from least to largest (as binary numbers).
If there are no possible configurations, output a single line with the single word `IMPOSSIBLE'
SAMPLE OUTPUT (file lamps.out)
0000000000 0101010101 0110110110
In this case, there are three possible final configurations:
- All lamps are OFF
- Lamps 1, 4, 7, 10 are OFF and lamps 2, 3, 5, 6, 8, 9 are ON.
- Lamps 1, 3, 5, 7, 9 are OFF and lamps 2, 4, 6, 8, 10 are ON.
题目大意就是给出灯的个数，和对灯的操作规则，和经过C步操作后亮着的部分灯和熄灭的部分灯，求灯最后的可能情况，如果没有输出`IMPOSSIBLE'
- 按钮1：当按下此按钮，将改变所有的灯：本来亮着的灯就熄灭，本来是关着的灯被点亮。
- 按钮2：当按下此按钮，将改变所有奇数号的灯。
- 按钮3：当按下此按钮，将改变所有偶数号的灯。
- 按钮4：当按下此按钮，将改变所有序号是3*K+1(K>=0)的灯。例如：1,4,7...
注意到当每对一个操作实施两次和没实施效果一样，故可将大于4的C减小到3或4；
然后从开始进行搜索没一盏灯进行一步操作后的可能情况，知道进行了C步操作，如果满足题目条件则保存数据再搜索下一组情况
/*
ID:shiryuw1
PROG:lamps
LANG:C++
*/
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int op[105];
int cl[105];
int n;
int c;
int clk=0;
int opk=0;
bool found=false;
struct prog{
int a1;
int a2;
int a3;
int a4;
bool str[105];
};
prog ans[10000];
int p=0;
bool istrue(bool* lap)
{
int i;
for(i=0;i<opk;i++)
{
if(lap[op[i]-1]!=true)
return false;
}
for(i=0;i<clk;i++)
{
if(lap[cl[i]-1]!=false)
return false;
}
return true;
}
void change(bool* lap,int k)
{
int st=0;
int ad=k;
if(k==4)
{
st=1;
ad=2;
}
int i;
for(i=st;i<n;i+=ad)
lap[i]=!lap[i];
}
void DFS(bool *lap,int k)
{
int i,j;
if(k/2==c/2)
{
if(istrue(lap))
{
for(j=0;j<n;j++)
ans[p].str[j]=lap[j];
for(j=0;j<25;j++)
{
ans[p].a1=ans[p].a1*2+ans[p].str[j];
}
for(j=25;j<50;j++)
{
ans[p].a2=ans[p].a2*2+ans[p].str[j];
}
for(j=50;j<75;j++)
{
ans[p].a3=ans[p].a3*2+ans[p].str[j];
}
for(j=75;j<100;j++)
{
ans[p].a4=ans[p].a4*2+ans[p].str[j];
}
p++;
found=true;
}
return ;
}
for(i=1;i<=4;i++)
{
bool tmp[105];
for(j=0;j<n;j++)
{
tmp[j]=lap[j];
}
change(tmp,i);
DFS(tmp,k+1);
}
}
int cmp(const void *a,const void *b)
{
if((*(prog *)a).a1!=(*(prog *)b).a1)
return (*(prog *)a).a1-(*(prog *)b).a1;
if((*(prog *)a).a2!=(*(prog *)b).a2)
return (*(prog *)a).a2-(*(prog *)b).a2;
if((*(prog *)a).a3!=(*(prog *)b).a3)
return (*(prog *)a).a3-(*(prog *)b).a3;
return (*(prog *)a).a4-(*(prog *)b).a4;
}
int main()
{
memset(ans,0,sizeof(ans));
bool lap[105];
freopen("lamps.in","r",stdin);
freopen("lamps.out","w",stdout);
memset(lap,true,sizeof(lap));
cin>>n;
cin>>c;
while(1)
{
cin>>op[opk];
if(op[opk]==-1)
break;
opk++;
}
while(1)
{
cin>>cl[clk];
if(cl[clk]==-1)
break;
clk++;
}
if(c>4)
{
if(c%2)
c=3;
else
c=4;
}
DFS(lap,0);
int i,j;
if(!found)
cout<<"IMPOSSIBLE"<<endl;
else
{
qsort(ans,p,sizeof(ans[0]),cmp);
for(i=0;i<p;i++)
{
if(i&&ans[i].a1==ans[i-1].a1&&ans[i].a2==ans[i-1].a2&&ans[i].a3==ans[i-1].a3&&ans[i].a4==ans[i-1].a4)
continue;
for(j=0;j<n;j++)
{
cout<<ans[i].str[j];
}
cout<<endl;
}
}
return 0;
}