USACO 2.2 Party Lamps

类型：枚举

描述：

 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 form3xK+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 Cis set to zero.

You are given the value of counter C (0 <= C <= 10000) and thefinal state of some of the lamps after some operations have been executed.Write a program to determine all the possible final configurations ofthe N lamps that are consistent with the given information, withoutrepetitions.

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 characterrepresents the state of lamp 1 and the last character represents thestate 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 fromleast to largest (as binary numbers).

If there are no possible configurations, output a single line withthe 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.

 

思路：1.有4种操作，操作的顺序与结果无关，两次同样的操作=无操作，于是可以枚举2⁴种操作；2.进一步的优化是，灯的状态可以压缩成6bit。

ps:请先自己实现，再参考标程。

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

#define MAXLAMP    6
#define LAMPMASK    ((1<<MAXLAMP)-1)

int nlamp;
int nswitch;
int ison;
int known;

int poss[1<<MAXLAMP];

int flip[4] = {
    LAMPMASK,        /* flip all lights */
    LAMPMASK & 0xAA,     /* flip odd lights */
    LAMPMASK & 0x55,    /* flip even lights */
    LAMPMASK & ((1<<(MAXLAMP-1))|(1<<(MAXLAMP-4)))    /* lights 1, 4 */
};

/*
 * Starting with current light state ``lights'', flip exactly n switches
 * with number >= i.
 */
void
search(int lights, int i, int n)
{
    if(n == 0) {
    if((lights & known) == ison)
        poss[lights] = 1;
    return;
    }

    for(; i<4; i++)
    search(lights ^ flip[i], i+1, n-1);
}

void
printseq(FILE *fout, int lights)
{
    int i;
    char s[100+1];

    for(i=0; i<nlamp; i++)
    s[i] = (lights & (1<<(MAXLAMP-1 - i%MAXLAMP))) ? '1' : '0';
    s[nlamp] = '\0';
    fprintf(fout, "%s\n", s);
}

void
main(void)
{
    FILE *fin, *fout;
    int a, i, impossible;

    fin = fopen("lamps.in", "r");
    fout = fopen("lamps.out", "w");
    assert(fin != NULL && fout != NULL);

    fscanf(fin, "%d %d", &nlamp, &nswitch);

    for(;;) {
    fscanf(fin, "%d", &a);
    if(a == -1)
        break;
    a = MAXLAMP-1 - (a-1) % MAXLAMP;
    ison |= 1<<a;
    known |= 1<<a;
    }

    for(;;) {
    fscanf(fin, "%d", &a);
    if(a == -1)
        break;
    a = MAXLAMP-1 - (a-1) % MAXLAMP;
    assert((ison & (1<<a)) == 0);
    known |= 1<<a;
    }

    if(nswitch > 4)
    if(nswitch%2 == 0)
        nswitch = 4;
    else
        nswitch = 3;

    for(; nswitch >= 0; nswitch -= 2)
        search(LAMPMASK, 0, nswitch);

    impossible = 1;
    for(i=0; i<(1<<MAXLAMP); i++) {
    if(poss[i]) {
        printseq(fout, i);
        impossible = 0;
    }
    }
    if(impossible)
    fprintf(fout, "IMPOSSIBLE\n");

    exit(0);
}