POJ-1903 Jurassic Remains

题目大意：

给出n个字符串，字符串仅由大写字母组成，问你用最多的字符串使得这些字符串里面的字符出现的总次数为偶数次

解题思路：

1.dfs+位运算

2.中途相遇法

第一种思路就是普通的搜索，因为数据规模不是非常大，所以用搜索加上位运算也是可以通过所有数据的。

第二种思路是中途相遇法，先考虑这n个字符串的前n/2个，把每个字符串的状态记录到map里面，再枚举后面的所有字符串的状态，在map中寻找是否存在这个状态。然后找到答案，时间复杂度是O(2^(n/2) * logn)或者更低

代码：

dfs+位运算

#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

string str;
int ans, status, num[26];
void dfs(int n, int flag, int cnt, int st) {
	if (n == 0) {
		if (cnt < ans) return;
		if (!flag) {ans = cnt; status = st;}
		return;
	}

	dfs(n-1, flag ^ num[n], cnt + 1, st | (1 << (n-1)));
	dfs(n-1, flag, cnt, st);
}
int main() {
	ios::sync_with_stdio(false); cin.tie(0);
	int n, len, pos;
	while (cin >> n) {
		vector<int> vec; vec.clear();
		memset(num, 0, sizeof(num));
		for (int i = 1; i <= n; ++i) {
			cin >> str;
			len = str.length();
			for (int j = 0; j < len; ++j) {
				num[i] ^= (1 << (str[j] - 'A'));
			}
		}
		ans = 0, status = 0, pos = 1;
		dfs(n, 0, 0, 0);
		cout << ans << endl;
		while (status) {
			if (status & 1) vec.push_back(pos);
			++pos; status >>= 1;
		}
		pos = vec.size();
		for (int i = 0; i < pos; ++i) {
			if (i) cout << " ";
			cout << vec[i];
		}
		cout << endl;
	}
	return 0;
}

中途相遇法

#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

string str;
vector<int> vec;
map<int, int> mp;
map<int, int>::iterator it;

int num[26];

inline int bitCount(int st) {
	int cnt = 0;
	while (st) {
		if (st & 1) ++cnt;
		st >>= 1;
	}
	return cnt;
}
int main() {
	ios::sync_with_stdio(false); cin.tie(0);
	int n, len, pos;
	while (cin >> n) {
		vec.clear(); mp.clear();
		int ans, res, status, half;
		memset(num, 0, sizeof(num));
		for (int i = 0; i < n; ++i) {
			cin >> str;
			len = str.length();
			for (int j = 0; j < len; ++j) {
				num[i] ^= (1 << (str[j] - 'A'));
			}
		}

		half = (1 << (n >> 1));
		for (int i = 0; i < half; ++i) {
			int flag = 0;
			for (int j = 0; j < (n >> 1); ++j) {
				if (i & (1 << j)) flag ^= num[j];
			}

			it = mp.find(flag);
			if (it == mp.end()) {
				mp[flag] = i;
			} else {
				if (bitCount(it -> second) < bitCount(i)) {
					it -> second = i;
				}
			}
		}

		ans = 0; res = 0;
		status = (1 << (n - (n >> 1)));
		for (int i = 0; i < status; ++i) {
			int flag = 0;
			for (int j = n >> 1; j < n; ++j) {
				if (i & (1 << (j - (n >> 1)))) {
					flag ^= num[j];
				}
			}

			it = mp.find(flag);
			if (it != mp.end()) {
				half = i << (n >> 1);
				int cnt = bitCount(half + it -> second);
				if (ans < cnt) {
					ans = cnt;
					res = half + it -> second;
				}
			}
		}
		cout << ans << endl;
		ans = 1;
		while (res) {
			if (res & 1) vec.push_back(ans);
			++ans;
			res >>= 1;
		}
		ans = vec.size();
		for (int i = 0; i < ans; ++i) {
			if (i) cout << " ";
			cout << vec[i];
		}
		cout << endl;
	}
	return 0;
}