返回顶部

模板 - 树状数组套主席树

实际上并不是主席树……只是有点像。

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

#define mid (l+r>>1)

//离散化之后的值是大一点的
const int MAXN = 200000 + 5;
int T[MAXN], tcnt;
//主席树每次修改稳定增加18个新节点
//每个元素加进来,修改的主席树由树状数组的复杂可以另外算,应该是至多增加100个节点,防止翻车直接330算了
int cnt[MAXN * 330], Lson[MAXN * 330], Rson[MAXN * 330];

int a[MAXN], val[MAXN], cn;

inline int Build(int l, int r) {
    int rt = ++tcnt;
    cnt[rt] = 0;
    if(l < r) {
        Lson[rt] = Build(l, mid);
        Rson[rt] = Build(mid + 1, r);
    }
    return rt;
}

inline void _Update(int &rt, int l, int r, int x, int d) {
    if(!rt)
        rt = ++tcnt;
    cnt[rt] += d;
    if(l < r) {
        if(x <= mid)
            _Update(Lson[rt], l, mid, x, d);
        else
            _Update(Rson[rt], mid + 1, r, x, d);
    }
}

//把x位置的数加入主席树森林
inline void Update1(int x, int v) {
    for(int i = x; i <= cn; i += (i & -i))
        _Update(T[i], 1, cn, a[x], 1);
}

//把x位置改成v
inline void Update2(int x, int v) {
    for(int i = x; i <= cn; i += (i & -i))
        _Update(T[i], 1, cn, a[x], -1);
    a[x] = v;
    for(int i = x; i <= cn; i += (i & -i))
        _Update(T[i], 1, cn, a[x], 1);
}

const int MAXLOGN = 80;
int Lroot[MAXLOGN], Rroot[MAXLOGN], ltop, rtop;

//[l,r]第k大
inline int Query(int l, int r, int k) {
    --l, ltop = rtop = 0;
    for(int i = l; i >= 1; i -= (i & -i))
        Lroot[++ltop] = T[i];
    for(int i = r; i >= 1; i -= (i & -i))
        Rroot[++rtop] = T[i];
    int LL = 1, RR = cn, MID;
    while(1) {
        MID = LL + RR >> 1;

        int sum = 0;
        for(int i = 1; i <= ltop; ++i)
            sum -= cnt[Lson[Lroot[i]]];
        for(int i = 1; i <= rtop; ++i)
            sum += cnt[Lson[Rroot[i]]];
        if(LL == MID) {
            if(sum >= k)
                return val[LL];
            return val[RR];
        }
        if(k <= sum) {
            //在左儿子
            for(int i = 1; i <= ltop; ++i)
                Lroot[i] = Lson[Lroot[i]];
            for(int i = 1; i <= rtop; ++i)
                Rroot[i] = Lson[Rroot[i]];
            RR = MID;
        } else {
            //在右儿子
            for(int i = 1; i <= ltop; ++i)
                Lroot[i] = Rson[Lroot[i]];
            for(int i = 1; i <= rtop; ++i)
                Rroot[i] = Rson[Rroot[i]];
            k -= sum;
            LL = MID + 1;
        }
    }
}

struct Q {
    char type;
    int l, r, x;
} q[MAXN];


int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    /*int c=100000;//n的上界
    int cntlog=0;
    for(int x=1;x<=c;++x){
        for(int i = x; i <= 200000; i += (i & -i))//cn的上界
            cntlog+=18;
    }
    printf("cntlog=%d\n",cntlog/200000);*/

    int n, m;
    scanf("%d%d", &n, &m);
    cn = 0;
    for(int i = 1; i <= n ; ++i) {
        scanf("%d", &a[i]);
        val[++cn] = a[i];
    }


    for(int i = 1; i <= m; ++i) {
        char s[2];
        scanf("%s", s);
        if(s[0] == 'Q') {
            q[i].type = 'Q';
            scanf("%d%d%d", &q[i].l, &q[i].r, &q[i].x);
        } else {
            q[i].type = 'C';
            scanf("%d%d", &q[i].l, &q[i].x);
            val[++cn] = q[i].x;
        }
    }

    sort(val + 1, val + 1 + cn);
    cn = unique(val + 1, val + 1 + cn) - (val + 1);

    for(int i = 1; i <= n; ++i)
        a[i] = lower_bound(val + 1, val + 1 + cn, a[i]) - val;

    for(int i = 1; i <= m; ++i)
        if(q[i].type == 'C')
            q[i].x = lower_bound(val + 1, val + 1 + cn, q[i].x) - val;

    tcnt = 0;
    T[0] = Build(1, cn);
    for(int i = 1; i <= n ; ++i)
        Update1(i, a[i]);
    for(int i = 1; i <= m; ++i) {
        if(q[i].type == 'Q')
            printf("%d\n", Query(q[i].l, q[i].r, q[i].x));
        else
            Update2(q[i].l, q[i].x);
    }
    return 0;
}
posted @ 2019-08-27 13:16  Inko  阅读(...)  评论(...编辑  收藏