题目链接

http://poj.org/problem?id=2965

分析

用二进制表示开关状态，暴力枚举每个位置是否操作即可

AC代码

#include <cstdio>
#include <cstdlib>
#include <iostream>

using namespace std;

const int maxn = 5;

int h[maxn], cnt;
char s[maxn];
pair<int, int> ans[maxn * maxn];

void dfs(int x, int y) {
	if (x == 4) {
		for (int i = 0; i < 4; ++i)
			if (h[i] != 15) return;
		printf("%d\n", cnt);
		for (int i = 1; i <= cnt; ++i)
			printf("%d %d\n", ans[i].first, ans[i].second);
		exit(0);
	}
	int nx = x, ny = y + 1;
	if (ny == 4) ++nx, ny = 0;
	dfs(nx, ny);
	h[x] ^= 15;
	for (int i = 0; i < 4; ++i)
		if (i != x) h[i] ^= 1 << y;
	++cnt;
	ans[cnt].first = x + 1, ans[cnt].second = y + 1;
	dfs(nx, ny);
	h[x] ^= 15;
	for (int i = 0; i < 4; ++i)
		if (i != x) h[i] ^= 1 << y;
	--cnt;
}

int main() {
	for (int i = 0; i < 4; ++i) {
		scanf("%s", s);
		for (int j = 0; j < 4; ++j)
			if (s[j] == '-') h[i] |= 1 << j;
	}
	dfs(0, 0);
	return 0;
}