題目鏈接

洛谷
bzoj

題解

整體二分

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;

inline int gi() {
    RG int x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-'0', c = getchar();
    return f ? -x : x;
}
const int N = 50010;
int n, m;

#define ls (rt<<1)
#define rs (rt<<1|1)

LL t[N<<2], z[N<<2];
inline void pushdown(int rt, int l, int r) {
    if (z[rt]) {
        int mid = (l + r) >> 1;
        t[ls] += (mid-l+1)*z[rt]; t[rs] += (r-mid)*z[rt];
        z[ls] += z[rt]; z[rs] += z[rt];
        z[rt] = 0;
    }
    return;
}
inline void update(int rt, int l, int r, int L, int R, int k) {
    if (L <= l && r <= R) {
        z[rt] += k;
        t[rt] += k*(r-l+1);
        return ;
    }
    int mid = (l + r) >> 1;
    pushdown(rt, l, r);
    if (L <= mid)
        update(ls, l, mid, L, R, k);
    if (R > mid)
        update(rs, mid+1, r, L, R, k);
    t[rt] = t[ls]+t[rs];
    return ;
}

inline LL query(int rt, int l, int r, int L, int R) {
    if (L <= l && r <= R) return t[rt];
    int mid = (l + r) >> 1;
    LL s = 0;
    pushdown(rt, l, r);
    if (L <= mid)
        s = query(ls, l, mid, L, R);
    if (R > mid)
        s += query(rs, mid+1, r, L, R);
    t[rt] = t[ls]+t[rs];
    return s;
}
struct Question {
    int op, a, b, id;
    LL c;
}q[N], lq[N], rq[N];
LL ans[N];

void div(int l, int r, int st, int ed) {
    if (st > ed) return ;
    if (l == r) {
        for (int i = st; i <= ed; i++)
            if (q[i].op == 2)
                ans[q[i].id] = l;
        return ;
    }
    int mid = (l + r) >> 1, lt = 0, rt = 0;
    for (int i = st; i <= ed; i++) {
        if (q[i].op == 1) {
            if (q[i].c <= mid)
                lq[++lt] = q[i];
            else update(1, 1, n, q[i].a, q[i].b, 1), rq[++rt] = q[i];
        }
        else {
            LL s = query(1, 1, n, q[i].a, q[i].b);
            if (s >= q[i].c) rq[++rt] = q[i];
            else q[i].c -= s, lq[++lt] = q[i];
        }
    }
    for (int i = st; i <= ed; i++)
        if (q[i].op == 1 && q[i].c > mid) update(1, 1, n, q[i].a, q[i].b, -1);
    for (int i = 1; i <= lt; i++)
        q[st+i-1] = lq[i];
    for (int i = 1; i <= rt; i++)
        q[st+lt+i-1] = rq[i];
    div(l, mid, st, st+lt-1); div(mid+1, r, st+lt, ed);
    return ;
}

int main() {
    n = gi(); m = gi();
    int k = 0;
    for (int i = 1; i <= m; i++) {
        q[i].op = gi(), q[i].a = gi(), q[i].b = gi();
        scanf("%lld", &q[i].c);
        q[i].id = (q[i].op == 2) ? ++k : 0;
    }
    div(-n, n, 1, m);
    for (int i = 1; i <= k; i++)
        printf("%lld\n", ans[i]);
    return 0;
}
 

洛谷 P3332 BZOJ 3110 [ZJOI2013]K大數查詢