cdq分治+樹狀數組

很久沒寫cdq分治了，大概有兩年了？

求答案最小字典序有個小trick，倒著求一個f[i]表示以i開始最長的序列長度，最後輸出的時候從前往後掃f，第一個等於答案且滿足前面已經選定的點帶來的限制的點就可以添加到答案序列裏來。

cdq分治求解中先處理後半部分，再計算後半部分的貢獻，再計算前半部分。註意，計算貢獻需要按某一維排序，然後類似歸並地處理，用bit來取max，但遞歸前半部分時需要恢復序列按id的增序。

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 5e4 + 100
 
;

struct Point {
    int x, y;
    int id;
} p[MAXN];
int f[MAXN];

bool cmp(const Point&a, const Point&b) {
    return a.id < b.id;
}
bool cmp2(const Point&a, const Point&b) {
    return a.x < b.x || (a.x == b.x && (a.y > b.y || (a.y == b.y && a.id < b.id)));
}

 
int b[MAXN * 2];
void upd(int p, int v) {
    //b[p] = v;
    //return ;
    for (; p < MAXN * 2; b[p] = max(b[p], v), p += p&-p);
}
void clr(int p) {
    //b[p] = 0;
    //return ;
    for (; p < MAXN * 2; b[p] = 0, p += p&-p);
}
int qry(int p) {
    int r = 0;
    //for (int i = 1; i <= p; ++i)
    
 
//    r = max(r, b[i]);
    //return r;
    for (; 0 < p; r = max(r, b[p]), p -= p&-p);
    return r;
}

void cdq(int l, int r) {
    if (l == r) return ;
    int m = (l + r) >> 1;
    cdq(m + 1, r);
    sort(p + m + 1, p + r + 1, cmp2);
    sort(p + l, p + m + 1, cmp2);
    for (int i = m, j = r; l <= i; --i) {
        while (m < j && p[i].x <= p[j].x) upd(p[j].y, f[p[j].id]), --j;
        f[p[i].id] = max(f[p[i].id], 1 + qry(p[i].y));
    }
    for (int j = r; m < j; --j)
        clr(p[j].y);
    sort(p + l, p + m + 1, cmp);
    cdq(l, m);
}

int main() {
#ifdef lol
    freopen("code.in", "r", stdin);
    freopen("code.out", "w", stdout);
#endif

    int n;
    while (scanf("%d", &n) != EOF) {
        int c = 0, *t = new int[n * 2];
        for (int i = 1; i <= n; ++i)
            scanf("%d", &p[i].y), t[c++] = p[i].y;
        for (int i = 1; i <= n; ++i)
            scanf("%d", &p[i].x), t[c++] = p[i].x;
        sort(t, t + c);
        for (int i = 1; i <= n; ++i) {
            p[i].x = lower_bound(t, t + c, p[i].x) - t + 1;
            p[i].y = lower_bound(t, t + c, p[i].y) - t + 1;
            p[i].id = i;
            f[i] = 1;
        }
        delete t;
        cdq(1, n);
        sort(p + 1, p + n + 1, cmp);
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            ans = max(ans, f[i]);
        }
        printf("%d\n", ans);
        for (int i = 1, j = 0; i <= n; ++i)
            if (f[i] == ans && (j == 0 || (p[j].x <= p[i].x && p[i].y <= p[j].y))) {
                printf("%d%c", i, " \n"[(--ans) == 0]);
                j = i;
            }
    }
    return 0;
}

HDU - 5324 Boring Class