POJ 1176 Party Lamps (DFS)

题目链接：POJ 1176

题意：

对于一串彩灯，提供四种改变彩灯状态（ON<=>OFF）的操作：a.改变所有彩灯状态；b.改变奇数彩灯状态；c.改变偶数彩灯状态；d.改变3k+1号彩灯状态（1,4,7,10...）。

 给定彩灯数目，操作次数，和对于某几个彩灯必须为ON、某几个彩灯必须为OFF的要求，问经过给定次数的操作，最终能达到的满足要求的状态有多少种，输出所有满足要求的彩灯状态（按二进制字符升序输出）。
CODE：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>

using namespace std;

const int maxn = 110;

char s[maxn*10][maxn];
int N, C, ontot, offtot, on[maxn], off[maxn], t, tot;
bool a[maxn];//a[i]=true:on a[i]=false:off
int vis[maxn*10];

int valid()
{
for (int i = 0; i < ontot; i++)
if (a[on[i]] == false) return 0;
for (int i = 0; i < offtot; i++)
if (a[off[i]] == true) return 0;
return 1;
}

void dfs(int k)
{
if (k > C) return;
if (k == C) {
if (valid()) {
s[tot][0] = '0';
for (int i = 1; i <= N; i++) {
if (a[i]) s[tot][i] = '1';
else s[tot][i] = '0';
}
s[tot][N + 1] = '\0';
tot++;
}
}
else {
for (int i = 1; i <= 4; i++) {
if (i == 1) {
for (int j = 1; j <= N; j++)
a[j] = !a[j];
dfs(k + 1);
for (int j = 1; j <= N; j++)
a[j] = !a[j];
}
else if(i == 2) {
for (int j = 1; j <= N; j += 2)
a[j] = !a[j];
dfs(k + 1);
for (int j = 1; j <= N; j += 2)
a[j] = !a[j];
}
else if (i == 3) {
for (int j = 2; j <= N; j += 2)
a[j] = !a[j];
dfs(k + 1);
for (int j = 2; j <= N; j += 2)
a[j] = !a[j];
}
else {
for (int j = 1; j <= N; j += 3)
a[j] = !a[j];
dfs(k + 1);
for (int j = 1; j <= N; j += 3)
a[j] = !a[j];
}
}
}
}

int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif
while (~scanf("%d%d", &N, &C)) {

if (C > 4) {//700K 16MS
C = C % 4 ;
}

/*
if (C>4)//700K 0MS
{
C %= 2;
if (C == 1) C = 3;
else C = 4;
}
*/
for (int i = 1; i <= N; i++)
a[i] = true;
ontot = offtot = tot= 0;
while (1) {
scanf("%d", &t);
if (t == -1) break;
on[ontot++] = t;
}
while (1) {
scanf("%d", &t);
if (t == -1) break;
off[offtot++] = t;
}
dfs(0);
//排序
for (int i = 0; i < tot; i++) {
int k = i;
for (int j = i + 1; j < tot; j++) {
if (strcmp(s[j]+1, s[k]+1) < 0) k = j;
}
if (k != i) {
char tmp[maxn];
strcpy(tmp, s[i]);
strcpy(s[i], s[k]);
strcpy(s[k], tmp);
}
}
//标记重复
memset(vis, 0, sizeof(vis));
for (int i = 0; i < tot; i++) {
if (vis[i]) continue;
for (int j = i + 1; j < tot; j++) {
if (strcmp(s[i], s[j]) == 0) vis[j] = 1;
//else break;
}
}

for (int i = 0; i < tot; i++) {
if (!vis[i]) {
//for (int j = 1; j <= N; j++)
// putchar(s[i][j]);
//printf("\n");
puts(s[i] + 1);
}
}
}
return 0;
}