UVaLive 2965 Jurassic Remains (状态压缩)

题意：给定 n 个大写字母组成的字符串，选择尽量多的串，使得大写字母都能出现偶数次。

析：由于n比较小，我们可以枚举前n/2的所有组合，然后再从后面查找。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <unordered_map>

#include <unordered_set>

#define debug() puts("++++");

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1LL << 60;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 24 + 5;

const int mod = 2000;

const int dr[] = {-1, 1, 0, 0};

const int dc[] = {0, 0, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

map<int, int> mp;

int a[maxn];

inline int bitcount(int x){ return x ? bitcount(x>>1) + (x&1) : 0; }

int main(){

    while(scanf("%d", &n) == 1){

        string s;

        for(int i = 0; i < n; ++i){

            cin >> s;

            a[i] = 0;

            for(int j = 0; j < s.size(); ++j)  a[i] ^= 1 << s[j] - 'A';

        }

        mp.clear();

        int n1 = n / 2, n2 = n - n1;

        for(int i = 0; i < (1<<n1); ++i){

            int tmp = 0;

            for(int j = 0; j < n1; ++j)  if(i & (1<<j)) tmp ^= a[j];

            if(!mp.count(tmp) || bitcount(mp[tmp]) < bitcount(i))  mp[tmp] = i;

        }

        int ans = 0;

        for(int i = 0; i < (1<<n2); ++i){

            int tmp = 0;

            for(int j = 0; j < n2; ++j)  if(i & (1<<j)) tmp ^= a[n1+j];

            if(mp.count(tmp) && bitcount(mp[tmp]) + bitcount(i) > bitcount(ans))  ans = mp[tmp] ^ (i<<n1);

        }

        printf("%d\n", bitcount(ans));

        for(int i = 0; i < n; ++i)  if(ans & (1<<i)) printf("%d ", i+1);

        printf("\n");

    }

    return 0;

}