BZOJ3196——二逼平衡树

1、题目大意：给你一个序列，有5种操作，都有什么呢。。。
1> 区间第k小 这个直接用二分+树套树做
2> 区间小于k的有多少 这个直接用树套树做
3> 单点修改 这个直接用树套树做
4> 区间内k的前驱 这个就是1和2操作的合并，就是查询k的排名，然后就是知道他的前驱的排名，然后第k小

5> 区间内k的后继 这个和4同理

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;struct Node{    Node *ch[2];    int num, cnt, v, r;    bool operator < (const Node& rhs) const{        return r < rhs.r;    }     inline int cmp(int x){        if(x == v) return -1;        if(x < v) return 0;        return 1;    }    inline void maintain(){        cnt = num;        if(ch[0]) cnt += ch[0] -> cnt;        if(ch[1]) cnt += ch[1] -> cnt;        return;    }} ft[5000000], *root[400000];int tot;int a[100000];inline void treap_rotate(Node* &o, int d){    Node *k = o -> ch[d ^ 1];     o -> ch[d ^ 1] = k  -> ch[d];    k -> ch[d] = o;    o -> maintain();    k -> maintain();    o = k;    return;}inline void treap_insert(Node* &o, int value){    if(!o){        o = &ft[tot ++];        o -> ch[0] = o -> ch[1] = NULL;        o -> num = 1;        o -> v = value;        o -> r = rand();    }    else{        int d = o -> cmp(value);        if(d == -1){            o -> num ++;        }        else{            treap_insert(o -> ch[d], value);            if(o -> ch[d] > o) treap_rotate(o, d ^ 1);        }    }    o -> maintain();}inline void treap_remove(Node* &o, int value){    if(!o) return;    int d = o -> cmp(value);    if(d == -1){        if(o -> num > 1) o -> num --;        else if(!o -> ch[0]) o = o -> ch[1];        else if(!o -> ch[1]) o = o -> ch[0];        else{            int d2;            if(o -> ch[0] > o -> ch[1]) d2 = 1;            else d2 = 0;            treap_rotate(o, d2);            treap_remove(o -> ch[d2], value);        }    }    else{        treap_remove(o -> ch[d], value);    }    if(o) o -> maintain();}inline int treap_lessk(Node* &o, int k){    if(!o) return 0;    int d = o -> cmp(k);    if(d == -1){        int ret = o -> num;        if(o -> ch[0]) ret += o -> ch[0] -> cnt;        return ret;    }    else if(d == 0){        return treap_lessk(o -> ch[d], k);    }    else{        int ss = o -> num;        if(o -> ch[0]) ss += o -> ch[0] -> cnt;        return treap_lessk(o -> ch[d], k) + ss;    }}inline int treap_find(Node* &o, int k){    if(!o) return 0;    int d = o -> cmp(k);    if(d == -1) return o -> num;    else return treap_find(o -> ch[d], k);}inline int segment_tree_find(int l, int r, int o, int x, int y, int k){    if(x <= l && r <= y){        return treap_find(root[o], k);    }    int mid = (l + r) / 2;    int ret = 0;    if(x <= mid) ret += segment_tree_find(l, mid, 2 * o, x, y, k);    if(y > mid) ret += segment_tree_find(mid + 1, r, 2 * o + 1, x, y, k);    return ret;}inline void segment_tree_add(int l, int r, int o, int x, int value){    treap_remove(root[o], a[x]);    treap_insert(root[o], value);    if(l == r){        a[x] = value;        return;    }    int mid = (l + r) / 2;    if(x <= mid) segment_tree_add(l, mid, 2 * o, x, value);    else segment_tree_add(mid + 1, r, 2 * o + 1, x, value);}inline int segment_tree_query_lessk(int l, int r, int o, int x, int y, int k){    if(x <= l && r <= y){        return treap_lessk(root[o], k);    }    int mid = (l + r) / 2;    int ret = 0;    if(x <= mid) ret += segment_tree_query_lessk(l, mid, 2 * o, x, y, k);    if(y > mid) ret += segment_tree_query_lessk(mid + 1, r, 2 * o + 1, x, y, k);    return ret;}inline int segment_tree_query_NO_k(int l, int r, int o, int x, int y, int k, int n){    int L = 0, R = 100000000;    while(L < R){        int mid = (L + R) / 2;        if(segment_tree_query_lessk(1, n, 1, x, y, mid) < k) L = mid + 1;        else R = mid;    }    return L;}int main(){    int n, m;    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; i ++){        int x;        scanf("%d", &x);        segment_tree_add(1, n, 1, i, x);    }    for(int i = 1; i <= m; i ++){        int op, x, y, z;        scanf("%d", &op);        if(op == 1){            scanf("%d%d%d", &x, &y, &z);            int ans = segment_tree_query_lessk(1, n, 1, x, y, z);            ans -= segment_tree_find(1, n, 1, x, y, z);            ans ++;            printf("%d\n", ans);        }        else if(op == 2){            scanf("%d%d%d", &x, &y, &z);            int ans = segment_tree_query_NO_k(1, n, 1, x, y, z, n);            printf("%d\n", ans);        }        else if(op == 3){            scanf("%d%d", &x, &y);            segment_tree_add(1, n, 1, x, y);        }        else if(op == 4){            scanf("%d%d%d", &x, &y, &z);            int k = segment_tree_query_lessk(1, n, 1, x, y, z);            k -= segment_tree_find(1, n, 1, x, y, z);            int ans = segment_tree_query_NO_k(1, n, 1, x, y, k, n);            printf("%d\n", ans);        }        else{            scanf("%d%d%d", &x, &y, &z);            int k = segment_tree_query_lessk(1, n, 1, x, y, z);            k ++;            int ans = segment_tree_query_NO_k(1, n, 1, x, y, k, n);            printf("%d\n", ans);        }    }    return 0;}