【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
", ans[i]); 
    return 0; 
}