usaco Party Lamps

不会做，看了解题报告的分析，豁然开朗，

每个按钮按2次和没按效果是一样的。所以每个按钮或者按或者不按，一共有2^4=16种状态。枚举每个按钮是否按下，然后生成结果，排序输出即可（注意判重）。

另外灯1和灯7,2和8,3和9...是一样的因此当N>=6时只需处理前6个,排序时转换为10进制数, 输出时反复输出前6个的状态.

code：

/*
ID: yueqiq
LANG: C++
TASK: lamps
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ls rt<<1
#define rs rt<<1|1
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("lamps.in","r",stdin)
#define writef freopen("lamps.out","w",stdout)
const double pi = acos(-1.0);
const int maxn = 50;
const int BigP = 99999999;
const int INF = 99999999;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
using namespace std;
struct node{
int state,step;
};
bool flag=true;
int res[50],num,tmp[7];
bool vis[500];
int N,C,light[101],off[101],lcnt,ocnt;
void bfs(){
node cur,tmp;
queue<node> q;
cur.state=63;
cur.step=0;
vis[63]=true;
q.push(cur);
while(!q.empty()){
tmp=q.front();q.pop();
if(tmp.step>C || tmp.step > 4) return;
res[++num]=tmp.state;
cur.state=tmp.state^63;
cur.step=tmp.step+1;
if(!vis[cur.state]){
q.push(cur);
vis[cur.state]=true;
}
cur.state=tmp.state^42;
if(!vis[cur.state]){
q.push(cur);
vis[cur.state]=true;
}
cur.state=tmp.state^21;
if(!vis[cur.state]){
q.push(cur);
vis[cur.state]=true;
}
cur.state=tmp.state^36;
if(!vis[cur.state]){
q.push(cur);
vis[cur.state]=true;
}
}
return;
}
void print(int sta){
int k=6,t;SET(tmp,0);
while(sta){
t=sta%2;
tmp[k--]=t;
sta/=2;
}
FF(i,lcnt){
light[i]=(light[i]-1)%6+1;
if(tmp[light[i]]!=1) return;
}
FF(i,ocnt){
off[i]=(off[i]-1)%6+1;
if(tmp[off[i]]==1) return;
}
int pos=1;
FOR(i,1,N){
printf("%d",tmp[pos++]);
if(pos>6) pos=1;
}
flag=false;
LN;
return ;
}
int main (){
readf;
writef;
int tmp;
SD(N);SD(C);
while(~SD(light[lcnt])&&light[lcnt]!=-1){
lcnt++;
}
while(~SD(off[ocnt])&&off[ocnt]!=-1){
ocnt++;
}
bfs();
sort(res+1,res+num+1);
FOR(i,1,num){
print(res[i]);
}
if(flag) puts("IMPOSSIBLE");
return 0;
}