USACO 2.2 Party Lamps

题目：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:NLine 2:Final value of CLine 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)

101-17 -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)

000000000001010101010110110110

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.

分析：

一、 开关1以1为周期，开关2和3以2为周期，开关4以3为周期，所以灯是以6=gcd（1，2，3）为周期进行变化的，可仅仅考虑6盏灯的情况。

二、每个开关按两次=没按，所以可将C不断减2直到小于5，得到有效按动开关的总次数。

light[k] 记录了从初始状态63（”111111“）经过c步能否到k，能则为1，不能则为0。 用深搜。这个数组最多64个元素（000000~111111）。

mark[6]记录题目中限制条件，因为6位一循环，7和1是一样的，8和2是一样的，等等。。

符合条件的字符串应该满足： light[k] == 1 并且 满足题目要求的限制条件。

最后，将字符串排序后输出。。

写的比较挫，因为最后我需要把满足条件字符串存起来，然后排序。 要是能不排序，直接输出，就好了。要是有人有方法，请告诉我。。。

代码：

/*ID: gjj50201LANG: C++TASK: lamps*/#include<stdio.h>#include<iostream>#include<string.h>#include<cmath>#include<vector>#include<algorithm>#define b1 63#define b2 21#define b3 42#define b4 9using namespace std;int N,C;int light[64];int mark[6];vector <string> ans;//搜索c步能达到的所有可能状态void dfs(int k, int steps){if(!steps){light[k] = 1;return;}int t;t = k^b1; dfs(t,steps-1);t = k^b2; dfs(t,steps-1);t = k^b3; dfs(t,steps-1);t = k^b4; dfs(t,steps-1);}//检查是否满足题目要求的限制条件int ok(int k){for(int i=0;i<6;i++)if( mark[i]!=-1 && (k&(1<<i)) != (mark[i]<<i))return 0;return 1;}//扩展成需要的N位串string convert(int k){int ans[6];//把十进制转化成二进制，并且相应位子上的数字存起来for(int i=0;i<6;i++)        ans[i] = (k & (1<<i)) / (1<<i);               string result = "";for(int i = 0; i<N ;i++)result += ans[i%6] + '0';    return result;}int main(){freopen("lamps.in","r",stdin);    freopen("lamps.out","w",stdout);    cin>>N>>C;    while(C>4) C-=2;    for(int i=0;i<6;i++)    mark[i] = -1;        int tmp;    cin>>tmp;    while(tmp!=-1) {mark[(tmp-1)%6] = 1; cin>>tmp;}        cin>>tmp;    while(tmp!=-1) {mark[(tmp-1)%6] = 0; cin>>tmp;}    dfs(63,C);    for(int i=0;i<64;i++)    if(light[i] && ok(i))    ans.push_back(convert(i));        if(!ans.size())    cout<<"IMPOSSIBLE"<<endl;    else{        sort(ans.begin(),ans.end());        vector<string>::iterator it;        for(it = ans.begin();it!=ans.end();it++)            cout<<*it<<endl;    }    return 0;}