【BZOJ3110】【LG3332】[ZJOI2013]K大数查询

【BZOJ3110】【LG3332】[ZJOI2013]K大数查询

题面

洛谷

BZOJ

题解

和普通的整体分治差不多

用线段树维护一下每个查询区间内大于每次二分的值\(mid\)的值即可

然后再按套路做就行了

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (ch != '-' && (ch > '9' || ch < '0')) ch = getchar();
    if (ch == '-') w = -1 , ch = getchar();
    while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
    return w * data;
}
typedef long long ll;
const int MAX_N = 50005;
#define lson (o << 1)
#define rson (o << 1 | 1)
struct SGT {
    bool clr; int tag; ll val;
    void clear() { clr = 0, tag = val = 0; }
} t[MAX_N << 2];
int N, M, ans[MAX_N];
void puttag(int o, int l, int r, int w) {
    t[o].val += 1ll * w * (r - l + 1);
    t[o].tag += w;
}
void clear(int o) {
    if (t[o].clr) {
        t[lson].clear(), t[rson].clear();
        t[lson].clr = t[rson].clr = 1;
        t[o].clr = 0;
    }
}
void pushdown(int o, int l, int r) {
    if (l == r) return ;
    int mid = (l + r) >> 1;
    clear(o);
    puttag(lson, l, mid, t[o].tag);
    puttag(rson, mid + 1, r, t[o].tag);
    t[o].tag = 0;
}
void modify(int o, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return (void)puttag(o, l, r, 1);
    pushdown(o, l, r);
    int mid = (l + r) >> 1;
    if (ql <= mid) modify(lson, l, mid, ql, qr);
    if (qr > mid) modify(rson, mid + 1, r, ql, qr);
    t[o].val = t[lson].val + t[rson].val;
}
ll query(int o, int l, int r, int ql, int qr) {
    pushdown(o, l, r);
    if (ql <= l && r <= qr) return t[o].val;
    int mid = (l + r) >> 1; ll res = 0;
    if (ql <= mid) res += query(lson, l, mid, ql, qr);
    if (qr > mid) res += query(rson, mid + 1, r, ql, qr);
    return res;
}
struct Query { int op, x, y; ll z; } q[MAX_N], lq[MAX_N], rq[MAX_N];
void Div(int lval, int rval, int st, int ed) {
    if (st > ed) return ;
    if (lval == rval) {
        for (int i = st; i <= ed; i++)
            if (q[i].op != 0) ans[q[i].op] = lval;
        return ;
    }
    int mid = (lval + rval) >> 1;
    int lt = 0, rt = 0;
    for (int i = st; i <= ed; i++) {
        if (q[i].op == 0) {
            if (q[i].z <= mid) lq[++lt] = q[i];
            else {
                rq[++rt] = q[i];
                modify(1, 1, N, q[i].x, q[i].y);
            }
        } else {
            ll res = query(1, 1, N, q[i].x, q[i].y);
            if (res >= q[i].z) rq[++rt] = q[i];
            else q[i].z -= res, lq[++lt] = q[i];
        }
    }
    for (int i = 1; i <= lt; i++) q[i + st - 1] = lq[i];
    for (int i = 1; i <= rt; i++) q[i + lt + st - 1] = rq[i];
    t[1].clear(); t[1].clr = 1;
    Div(lval, mid, st, st + lt - 1);
    Div(mid + 1, rval, st + lt, ed);
}
int main () {
    int tot = 0;
    N = gi(), M = gi();
    for (int i = 1; i <= M; i++) {
        q[i].op = gi() - 1;
        q[i].x = gi(), q[i].y = gi(), q[i].z = gi();
        if (q[i].op) q[i].op = ++tot;
    }
    Div(-N, N, 1, M);
    for (int i = 1; i <= tot; i++) printf("%d\n", ans[i]);
    return 0;
}