「ZJOI2013」K大数查询

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「ZJOI2013」K大数查询

传送门
整体二分，修改的时候用线段树代替树状数组即可。
参考代码：

#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

typedef long long LL;
const int _ = 5e4 + 5;

int n, m, q, res[_]; LL sum[_ << 2]; int tag[_ << 2];
struct node { int opt, l, r; LL c; int id; } t[_], tt1[_], tt2[_];

inline int lc(int p) { return p << 1; }

inline int rc(int p) { return p << 1 | 1; }

inline void pushup(int p) { sum[p] = sum[lc(p)] + sum[rc(p)]; }

inline void f(int p, int l, int r, int v) { sum[p] += 1ll * v * (r - l + 1), tag[p] += v; }

inline void pushdown(int p, int l, int r, int mid)
{ if (tag[p]) f(lc(p), l, mid, tag[p]), f(rc(p), mid + 1, r, tag[p]), tag[p] = 0; }

inline void update(int ql, int qr, int v, int p = 1, int l = 1, int r = n) {
    if (ql <= l && r <= qr) return f(p, l, r, v);
    int mid = (l + r) >> 1;
    pushdown(p, l, r, mid);
    if (ql <= mid) update(ql, qr, v, lc(p), l, mid);
    if (qr > mid) update(ql, qr, v, rc(p), mid + 1, r);
    pushup(p);
}

inline LL query(int ql, int qr, int p = 1, int l = 1, int r = n) {
    if (ql <= l && r <= qr) return sum[p];
    int mid = (l + r) >> 1; LL res = 0;
    pushdown(p, l, r, mid);
    if (ql <= mid) res += query(ql, qr, lc(p), l, mid);
    if (qr > mid) res += query(ql, qr, rc(p), mid + 1, r);
    return res;
}

inline void solve(int ql, int qr, int l, int r) {
    if (ql > qr || l > r) return ;
    if (l == r) { for (rg int i = ql; i <= qr; ++i) if (t[i].opt == 2) res[t[i].id] = l; return ; }
    int mid = (l + r) >> 1, p1 = 0, p2 = 0;
    for (rg int i = ql; i <= qr; ++i) {
        if (t[i].opt == 1) {
            if (t[i].c <= mid) tt1[++p1] = t[i]; else update(t[i].l, t[i].r, 1), tt2[++p2] = t[i];
        } else {
            LL cnt = query(t[i].l, t[i].r);
            if (cnt < t[i].c) t[i].c -= cnt, tt1[++p1] = t[i]; else tt2[++p2] = t[i];
        }
    }
    for (rg int i = 1; i <= p2; ++i) if (tt2[i].opt == 1) update(tt2[i].l, tt2[i].r, -1);
    for (rg int i = 1; i <= p1; ++i) t[ql + i - 1] = tt1[i];
    for (rg int i = 1; i <= p2; ++i) t[ql + p1 + i - 1] = tt2[i];
    solve(ql, ql + p1 - 1, l, mid), solve(ql + p1, qr, mid + 1, r);
}

int main() {
    read(n), read(m);
    for (rg int i = 1; i <= m; ++i)
        read(t[i].opt), read(t[i].l), read(t[i].r), read(t[i].c), t[i].id = t[i].opt == 2 ? ++q : 0;
    solve(1, m, 0, n);
    for (rg int i = 1; i <= q; ++i) printf("%d\n", res[i]);
    return 0;
}

「ZJOI2013」K大数查询

标签：参考   query   long   传送门   cpp   ++i   return   res   include   

原文地址：https://www.cnblogs.com/zsbzsb/p/12231706.html