P3604 美好的每一天 题解

题目链接：美好的每一天

经典题，这种字符串重排以后回文，优先考虑异或操作。考虑对一个字符串状态压缩，每个字符表示为 \(1<<(c-97)\)，然后将它们异或起来就可以得到一个字符串的状态压缩情况。

考虑每一位异或情况，如果为偶数个单个字符则为 \(0\)，奇数则为 \(1\)，注意可以重排，那么当且仅当 \(1\) 的数量 \(<=1\) 时为回文串。而区间异或，我们可以转化为前缀异或。

注意到前缀异或为 \(pre[r]-pre[l-1]\)，所以我们查询的时候还需要加入 \(l-1\) 处的点，保证个数不漏。

考虑单点加入和删除的贡献情况，第一种就是和自身异或为 \(0\) 的，第二种则是异或为某个 \(1\)，即 \(2\) 的次方，由于至多 \(26\) 种，又观察到 \(n \le 6 \times 10^4\)，那么枚举就行了。注意下，删除的时候先删除自身，因为删除会去掉和自身异或为 \(0\) 的情况，所以要优先排除自身和自身异或，增加则先加入和其他异或的情况，再加入自身。跑个普通莫队预处理出前缀异或数组即可，每次加入或者删除为前缀异或数组里的某个值。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

#define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}


template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 6e4 + 10;
int pos[N];

struct Mo
{
    int l, r, id;

    bool operator<(const Mo& other) const
    {
        return pos[l] ^ pos[other.l] ? l < other.l : pos[l] & 1 ? r < other.r : r > other.r;
    }
} node[N];

int pre[N];
ll ans;
constexpr int MX = 1 << 26 | 1;
int cnt[MX];

inline void add(const int curr)
{
    ans += cnt[curr];
    cnt[curr]++;
    forn(i, 0, 25)ans += cnt[curr ^ 1 << i];
}

inline void del(const int curr)
{
    cnt[curr]--;
    ans -= cnt[curr];
    forn(i, 0, 25)ans -= cnt[curr ^ 1 << i];
}

int n, q;
char c;
ll res[N];

inline void solve()
{
    cin >> n >> q;
    const int siz = sqrt(n);
    forn(i, 1, n)cin >> c, pre[i] = 1 << c - 'a', pre[i] ^= pre[i - 1], pos[i] = (i - 1) / siz + 1;
    forn(i, 1, q)
    {
        auto& [l,r,id] = node[i];
        cin >> l >> r, id = i;
        l--;
    }
    sortArr(node, q);
    int l = 1, r = 0;
    forn(i, 1, q)
    {
        auto [L,R,id] = node[i];
        while (l < L)del(pre[l++]);
        while (l > L)add(pre[--l]);
        while (r < R)add(pre[++r]);
        while (r > R)del(pre[r--]);
        res[id] = ans;
    }
    forn(i, 1, q)cout << res[i] << endl;
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[时间复杂度为:\ O(26n\sqrt{q}) \]