BZOJ3196: Tyvj 1730 二逼平衡树 (线段树 + Treap 练习题)

线段树+Treap, 基本思路同ZOJ2112(区间kth):  http://www.cnblogs.com/usingnamespace/p/5152220.html

维护x在区间[s, t]的前驱只要利用线段树将该区间分为线段树上一些节点代表的区间, 在每个区间中求x的前驱取其中的

最大值

同理后继只需在[s, t]划分出的每个区间中求后继然后取最小值

代码如下(写法拙劣, 最开始少写了个else, TLE了, 最后过了但还是跑了10108 ms, 指针写法快一点吧)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef const int& ci;
const int maxn = 3000007, inf = 1e9;
int read() {
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
    return x * f;
}
int n, m, Nm, Pre, Suf, arr[maxn];

int root[maxn], sz, size[maxn], rnd[maxn], val[maxn], lc[maxn], rc[maxn], cnt[maxn];
void update(ci k) {
    size[k] = size[lc[k]] + size[rc[k]] + cnt[k];
}
void r_rot(int &k) {
    int t = lc[k]; lc[k] = rc[t]; rc[t] = k;
    size[t] = size[k]; update(k); k = t;
}
void l_rot(int &k) {
    int t = rc[k]; rc[k] = lc[t]; lc[t] = k;
    size[t] = size[k]; update(k); k = t;
}
void Insert(int &k, ci x) {
    if(k == 0) {
        size[k = ++sz] = cnt[k] = 1; val[k] = x;
        rnd[k] = rand(); lc[k] = rc[k] = 0; return;
    }
    size[k]++;
    if(x == val[k]) {cnt[k]++; return;}
    if(x < val[k]) {
        Insert(lc[k], x);
        if(rnd[lc[k]] < rnd[k]) r_rot(k);
    } else {
        Insert(rc[k], x);
        if(rnd[rc[k]] < rnd[k]) l_rot(k);
    }
} 
void Delete(int &k , ci x) {
    if(k == 0) return;
    if(x == val[k]) {
        if(cnt[k] > 1) {cnt[k]--, size[k]--; return;}
        if(lc[k] * rc[k] == 0) k = lc[k] + rc[k];
        else if(rnd[lc[k]] < rnd[rc[k]]) r_rot(k), Delete(k, x);
        else l_rot(k), Delete(k, x);
    } else if(x < val[k]) size[k]--, Delete(lc[k], x);
    else size[k]--, Delete(rc[k], x);
}
//求不大于num的数的个数 
void Find(ci k, ci num) {
    if(!k) return;
    if(num >= val[k]) {
        Nm += size[lc[k]] + cnt[k];
        Find(rc[k], num);
    } else Find(lc[k], num);
}
void pre(ci k, ci x) {
    if(!k) return;
    if(x > val[k]) {
        Pre = max(Pre, val[k]);
        pre(rc[k], x);
    } else pre(lc[k], x);
}
void suf(ci k, ci x) {
    if(!k) return;
    if(x < val[k]) {
        Suf = min(Suf, val[k]);
        suf(lc[k], x);
    } else suf(rc[k], x);
}

    
void insert(ci k, ci l, ci r, ci v, ci pos) {
    Insert(root[k], v);
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(pos <= mid) insert(k << 1, l, mid, v, pos);
    else insert(k << 1 | 1, mid + 1, r, v, pos);
}
void modify(ci k, ci l, ci r, ci v, ci pos) {
    Delete(root[k], arr[pos]);
    Insert(root[k], v);
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(pos <= mid) modify(k << 1, l, mid, v, pos);
    else modify(k << 1 | 1, mid + 1, r, v, pos);
}
//区间内排名 
void Rank(ci k, ci l, ci r, ci s, ci t, ci x) {
    if(l == s && t == r) {Find(root[k], x); return;}
    int mid = (l + r) >> 1;
    if(t <= mid) {Rank(k << 1, l, mid, s, t, x); return;}
    if(s > mid) {Rank(k << 1 | 1, mid + 1, r, s, t, x); return;}
    Rank(k << 1, l, mid, s, mid, x); Rank(k << 1 | 1, mid + 1, r, mid + 1, t, x);
}
//区间内前驱 
void Prev(ci k, ci l, ci r, ci s, ci t, ci x) {
    if(l == s && t == r) {pre(root[k], x); return;}
    int mid = (l + r) >> 1; 
    if(t <= mid) {Prev(k << 1, l, mid, s, t, x); return;}
    if(s > mid) {Prev(k << 1 | 1, mid + 1, r, s, t, x); return;}
    Prev(k << 1, l, mid, s, mid, x); Prev(k << 1 | 1, mid + 1, r, mid + 1, t, x);
}
//区间内后继 
void Suff(ci k, ci l, ci r, ci s, ci t, ci x) {
    if(l == s && t == r) {suf(root[k], x); return;}
    int mid = (l + r) >> 1;
    if(t <= mid) {Suff(k << 1, l, mid, s, t, x); return;}
    if(s > mid) {Suff(k << 1 | 1, mid + 1, r, s, t, x); return;}
    Suff(k << 1, l, mid, s, mid, x); Suff(k << 1 | 1, mid + 1, r, mid + 1, t, x);    
}
int main() {
    n = read(); m = read();
    for(int i = 1; i <= n; i++) {
        arr[i] = read();
        insert(1, 1, n, arr[i], i);
    }
    for(int i = 1; i <= m; i++) {
        int opt, x, y, z, l, r; opt = read();
        switch(opt) {
            case 1: 
                x = read(); y = read(); z = read();
                Nm = 0; Rank(1, 1, n, x, y, z - 1);
                printf("%d\n", Nm + 1); break;
            case 2: 
                x = read(); y = read(); z = read();
                l = 0, r = 100000000;
                while(l <= r) {
                    int mid = (l + r) >> 1; Nm = 0;
                    Rank(1, 1, n, x, y, mid);
                    if(Nm >= z) r = mid - 1;
                    else l = mid + 1;
                }
                printf("%d\n", l); break;
            case 3: 
                x = read(); y = read();
                modify(1, 1, n, y, x); 
                arr[x] = y; break;
            case 4: 
                x = read(); y = read(); z = read();
                Pre = 0; Prev(1, 1, n, x, y, z); 
                printf("%d\n", Pre); break;
            case 5: 
                x = read(); y = read(); z = read();
                Suf = inf; Suff(1, 1, n, x, y, z); 
                printf("%d\n", Suf); break;
        }
    }
    return 0;
}
posted @ 2016-01-26 15:48  using_namespace  阅读(184)  评论(0编辑  收藏  举报