HDU - 5845 Best Division dp + 字典树

HDU - 5845

dp[ i ] 表示分完前 i 段， 最多能分几段。

我们能得到一个n2的dp， 然后用字典树优化掉。

我用了一个multiset去维护删除， 但实际上因为dp值有单调性， 所有维护sz就够了。

换成c++卡内从卡过去的。

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<set>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int LOG = 28;

int n, up, len;
int a[N];
int dp[N];

void read() {
    int p, q;
    scanf("%d%d%d", &a[1], &p, &q);
    for(int i = 2; i <= n; i++) {
        a[i] = (1LL * a[i - 1] * p + q) % 268435456;
    }
    for(int i = 1; i <= n; i++) {
        a[i] ^= a[i - 1];
    }
}

int trietot, Rt;

struct Trie {
    int ch[2], mx;
} tr[N * LOG];

multiset<int> mulset[N * LOG];

inline int newNode() {
    trietot++;
    tr[trietot].ch[0] = 0;
    tr[trietot].ch[1] = 0;
    tr[trietot].mx = -1;
    mulset[trietot].clear();
    return trietot;
}

void ins(int x, int val) {
    int u = Rt;
    for(int i = LOG - 1; i >= 0; i--) {
        chkmax(tr[u].mx, val);
        int v = tr[u].ch[x >> i & 1];
        if(!v) tr[u].ch[x >> i & 1] = newNode();
        v = tr[u].ch[x >> i & 1];
        u = v;
    }
    mulset[u].insert(val);
    chkmax(tr[u].mx, val);
}

void del(int u, int x, int val, int d) {
    if(d == -1) {
        mulset[u].erase(mulset[u].lower_bound(val));
        if(SZ(mulset[u])) tr[u].mx = *mulset[u].rbegin();
        else tr[u].mx = -1;
        return;
    }
    del(tr[u].ch[x >> d & 1], x, val, d - 1);
    tr[u].mx = max(tr[tr[u].ch[0]].mx, tr[tr[u].ch[1]].mx);
}

int query(int x) {
    int ans = -1;
    bool limit = true;
    bool ban = false;
    int u = Rt;
    for(int i = LOG - 1; i >= 0; i--) {
        if(limit) {
            int sml = x >> i & 1;
            int big = sml ^ 1;
            int smlv = tr[u].ch[sml];
            int bigv = tr[u].ch[big];
            if(!(up >> i & 1)) {
                u = smlv;
            }
            else {
                if(tr[smlv].mx >= tr[bigv].mx) {
                    limit = false;
                    u = smlv;
                }
                else {
                    chkmax(ans, tr[smlv].mx);
                    u = bigv;
                }
            }
        }
        else {
            chkmax(ans, tr[u].mx);
            ban = true;
            break;
        }
    }
    if(!ban) chkmax(ans, tr[u].mx);
    return ans;
}

void init() {
    tr[0].mx = -1;
    trietot = 0;
    Rt = newNode();
}

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        init();
        scanf("%d%d%d", &n, &up, &len);
        read();
        ins(0, 0);
        for(int i = 1, j = 0; i <= n; i++) {
            while(j + len < i) {
                if(dp[j] != -1) del(Rt, a[j], dp[j], LOG - 1);
                j++;
            }
            dp[i] = query(a[i]);
            if(dp[i] != -1) {
                dp[i]++;
                ins(a[i], dp[i]);
            }
        }
        if(dp[n] != -1) printf("%d\n", dp[n]);
        else puts("0");
    }
    return 0;
}

/*
*/