HDU - 5324 Boring Class BIT套线段树

HDU - 5324 

直接树套树维护dp， 或者分治 + 树状数组维护。

好像内存卡得比较紧。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#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 = 5e4 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)1e9 + 7;
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());

int n;
int L[N], R[N];
int hsL[N], hsR[N];
int totL, totR;
int path[N];
int dp[N];
vector<int> ans;

struct Node {
    int id, ls, rs;
} a[N * 80];

int treeTot;

inline int newNode() {
    treeTot++;
    a[treeTot].id = 0;
    a[treeTot].ls = 0;
    a[treeTot].rs = 0;
    return treeTot;
}

void update(int p, int id, int l, int r, int &rt) {
    if(!rt) rt = newNode();
    if(dp[a[rt].id] < dp[id] || dp[a[rt].id] == dp[id] && id < a[rt].id) a[rt].id = id;
    if(l == r) return;
    int mid = l + r >> 1;
    if(p <= mid) update(p, id, l, mid, a[rt].ls);
    else update(p, id, mid + 1, r, a[rt].rs);
}

int query(int L, int R, int l, int r, int rt) {
    if(!rt) return 0;
    if(R < l || r < L || R < L) return 0;
    if(L <= l && r <= R) return a[rt].id;
    int mid = l + r >> 1;
    int id = query(L, R, l, mid, a[rt].ls);
    int rid = query(L, R, mid + 1, r, a[rt].rs);
    if(dp[id] < dp[rid] || dp[id] == dp[rid] && id > rid) id = rid;
    return id;
}


struct Bit {
    int a[N];
    void init() {
        for(int i = 1; i <= totL; i++) {
            a[i] = newNode();
        }
    }
    void modify(int x, int y, int id) {
        for(int i = x; i <= totL; i += i & -i) {
            update(y, id, 1, totR, a[i] );
        }
    }
    int query(int x, int y) {
        int id = 0;
        for(int i = x; i; i -= i & -i) {
            int rid = ::query(y, totR, 1, totR, a[i]);
            if(dp[id] < dp[rid] || dp[id] == dp[rid] && id > rid) id = rid;
        }
        return id;
    }
} bit;

void init() {
    totL = totR = treeTot =  0;
    ans.clear();
    for(int i = 1; i <= n; i++) {
        path[i] = 0;
    }
}

int main() {
    while(scanf("%d", &n) != EOF) {
        init();
        for(int i = 1; i <= n; i++) {
            scanf("%d", &L[i]);
            hsL[++totL] = L[i];
        }
        for(int i = 1; i <= n; i++) {
            scanf("%d", &R[i]);
            hsR[++totR] = R[i];
        }
        sort(hsL + 1, hsL + 1 + totL);
        sort(hsR + 1, hsR + 1 + totR);
        totL = unique(hsL + 1, hsL + 1 + totL) - hsL - 1;
        totR = unique(hsR + 1, hsR + 1 + totR) - hsR - 1;
        for(int i = 1; i <= n; i++) {
            L[i] = lower_bound(hsL + 1, hsL + 1 + totL, L[i]) - hsL;
            R[i] = lower_bound(hsR + 1, hsR + 1 + totR, R[i]) - hsR;
        }
        bit.init();
        int mx = 0;
        for(int i = n; i >= 1; i--) {
            int id = bit.query(L[i], R[i]);
            dp[i] = 1;
            if(id > 0) {
                dp[i] = dp[id] + 1;
                path[i] = id;
            }
            chkmax(mx, dp[i]);
            bit.modify(L[i], R[i], i);
        }
        int start = -1;
        for(int i = 1; i <= n; i++) {
            if(dp[i] == mx) {
                start = i;
                break;
            }
        }
        for(int i = start; i; i = path[i]) {
            ans.push_back(i);
        }
        printf("%d\n", mx);
        for(int i = 0; i < SZ(ans); i++) {
            printf("%d%c", ans[i], " \n"[i == SZ(ans) - 1]);
        }
    }
    return 0;
}

/*
*/