UVa 1326 - Jurassic Remains

题意

找出尽量多的串，使每个字母都出现偶数次

思路

因为N<=24，考虑状态压缩。而对于每一个状态，可以用1表示出现奇数，0表示出现偶数，再对一个串中的字母进行状态压缩。

但是如果一直^到最后，复杂度为O(2n)，TLE。考虑枚举一半的串的状态，然后在后一半中查找。因为只有两个相同的状态值异或才可以符合条件。

这样复杂度就变成O(1.414n)

代码

#include<bits/stdc++.h>
#define LL long long
#define lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
const int MAXN = 30 + 5;
const int INF = 0x3f3f3f3f;
using namespace std;
typedef pair<int, int> pii;
typedef vector<int> vei;
typedef vector<pair<int, int> >veii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
typedef priority_queue<pii, vector<pii>, greater<pii> >pquii;
char word[1000];
int num[MAXN];
map<int, int> mp;
int BitCount(int x)
{
    return x == 0 ? 0 : BitCount(x >> 1) + (x & 1);
}
int main()
{
    //freopen("input.txt", "r", stdin);
    int n, i, j;
    while (~scanf("%d", &n))
    {
        mp.clear();
        for (i = 0; i < n; i++)
        {
            scanf("%s", word);
            num[i] = 0;
            for (j = 0; word[j]; j++)
                num[i] ^= (1 << (word[j] - 'A'));
        }
        int n1 = n / 2, n2 = n - n1;
        for (int state = 0; state < (1 << n1); state++)
        {
            int cur = 0;
            for (i = 0; i < n; i++)
                if ((1 << i) & state) cur ^= num[i];
            if (!mp.count(cur) || BitCount(mp[cur]) < BitCount(state)) mp[cur] = state;
        }
        int ans = 0;
        for (int state = 0; state < (1 << n2); state++)
        {
            int cur = 0;
            for (int i = 0; i < n2; i++) if (state & (1 << i)) cur ^= num[n1 + i];
            if (mp.count(cur) && BitCount(ans) < BitCount(mp[cur]) + BitCount(state))
                ans = (state << n1) ^ mp[cur];
        }
        bool first = true;
        printf("%d\n", BitCount(ans));
        for (i = 0; i < n; i++)
            if ((1 << i) & ans)
            {
                if (first)
                    printf("%d", i + 1), first = false;
                else printf(" %d", i + 1);
            }
        puts("");
    }
    return 0;
}