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poj3468 A Simple Problem with Integers 指针版splay

时间:2015-07-03 17:26:17      阅读:119      评论:0      收藏:0      [点我收藏+]

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题目链接:poj 3468

#include <iostream>
#include <fstream>
#include <string>
#include <time.h>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if (c = getchar(), c == EOF) return 0;
    while (c != ‘-‘ && (c<‘0‘ || c>‘9‘)) c = getchar();
    sgn = (c == ‘-‘) ? -1 : 1;
    ret = (c == ‘-‘) ? 0 : (c - ‘0‘);
    while (c = getchar(), c >= ‘0‘&&c <= ‘9‘) ret = ret * 10 + (c - ‘0‘);
    ret *= sgn;
    return 1;
}
template <class T>
inline void pt(T x) {
    if (x <0) {
        putchar(‘-‘);
        x = -x;
    }
    if (x>9) pt(x / 10);
    putchar(x % 10 + ‘0‘);
}
typedef long long ll;
#define lsum (ch[0]->sum)
#define rsum (ch[1]->sum)
#define lkey (ch[0]->key)
#define rkey (ch[1]->key)
#define lsiz (ch[0]->size)
#define rsiz (ch[1]->size)
#define lson ch[0]
#define rson ch[1]
#define llazy (ch[0]->lazy)
#define rlazy (ch[1]->lazy)

const int N = 1e5 + 10;
const int MAXN = 50010;
struct Node *null;
struct Node{
    Node *ch[2], *fa;
    int size, cnt;//cnt表示当前这个点有一个节点,虚拟节点是0
    ll lazy, sum, key;
    inline void setc(Node *p, int d){//p是this的d儿子
        ch[d] = p;
        p->fa = this;
    }
    inline bool d(){//这个节点是右儿子
        return fa->ch[1] == this;
    }
    inline void push_up(){
        size = lsiz + rsiz + cnt;
        sum = lsum + rsum + key;
    }
    inline void push_down(){
        if (!lazy)return;
        if (lson != null){ lkey += lazy; lsum += lsiz * lazy; llazy += lazy; }
        if (rson != null){ rkey += lazy; rsum += rsiz * lazy; rlazy += lazy; }
        lazy = 0;
    }
    void clear(int _key){
        size = cnt = 1;
        sum = key = _key;
        ch[0] = ch[1] = fa = null;
        lazy = 0;
    }
    inline bool isroot(){
        return fa == null || this != fa->ch[0] && this != fa->ch[1];
    }
};
Node pool[MAXN * 15], *tail;
Node *bc[MAXN];
int bc_top;//内存回收
void init(){
    tail = pool;
    bc_top = 0;
    null = tail++;
    null->size = null->cnt = 0;
    null->ch[0] = null->ch[1] = null->fa = null;
}
inline void rotate(Node *x){
    Node *f = x->fa, *ff = x->fa->fa;
    f->push_down(); x->push_down();
    int c = x->d(), cc = f->d();
    f->setc(x->ch[!c], c);
    x->setc(f, !c);
    if (ff->ch[cc] == f)ff->setc(x, cc);
    else x->fa = ff;
    f->push_up();
}
inline void splay(Node* &root, Node* x, Node* goal){
    x->push_down();
    while (x->fa != goal){
        if (x->fa->fa == goal)rotate(x);
        else {
            bool f = x->fa->d();
            x->d() == f ? rotate(x->fa) : rotate(x);
            rotate(x);
        }
    }
    x->push_up();
    if (goal == null)root = x;
}
//找到r子树里面的最左边那个
Node* get_left(Node* r){
    Node* x = r;
    while (x->ch[0] != null)x = x->ch[0];
    return x;
}
//在root的树中删掉x
void erase(Node* &root, Node* x){
    splay(root, x, null);
    Node* t = root;
    if (t->ch[1] != null){
        root = t->ch[1];
        splay(root, get_left(t->ch[1]), null);
        root->setc(t->ch[0], 0);
    } 
    else root = root->ch[0];
    bc[bc_top++] = x;
    root->fa = null;
    if (root != null)root->push_up();
}
Node* newNode(int key){
    Node* p;
    if (bc_top)p = bc[--bc_top];
    else p = tail++;
    p->clear(key);
    return p;
}
Node* get_kth(Node* x, int k){//寻找x后面的第k个
    x->push_down();
    Node*u = x;
    while (u->ch[0]->size+1 != k)
    {
        if (u->ch[0]->size >= k)
            u = u->ch[0];
        else {
            k -= u->ch[0]->size + 1;
            u = u->ch[1];
        }
        u->push_down();
    }
    return u;
}
//插入一个值key
void insert(Node* &root, int key){
    if (root == null){
        root = newNode(key);
        return;
    }
    Node* now = root;
    Node* pre = root->fa;
    while (now != null){
        if (now->key == key){
            now->cnt++;
            splay(root, now, null);
            return;
        }
        pre = now;
        now = now->ch[key >= now->key];
    }
    Node *x = newNode(key);
    pre->setc(x, key >= pre->key);
    splay(root, x, null);
}
//删除一个值key
void erase(Node* &root, int key){
    Node* now = root;
    while (now->key != key){
        now = now->ch[key >= now->key];
    }
    now->cnt--;
    if (now->cnt == 0)erase(root, now);
    else splay(root, now, null);
}
void Travel(Node* r){
    if (r == null)return;
    Travel(r->ch[0]);
    bc[bc_top++] = r;
    Travel(r->ch[1]);
}
void CLEAR(Node* &root){
    Travel(root);
    root = null;
}
//查询小于等于val的个数
int query(Node *root, int val){
    int ans = 0;
    Node *x = root;
    while (x != null){
        if (val < x->key)x = x->ch[0];
        else{
            ans += x->ch[0]->size + x->cnt;
            x = x->ch[1];
        }
    }
    return ans;
}
Node* build(int l, int r, int *a){
    if (l > r)return null;
    int mid = (l + r) >> 1;
    Node* u = newNode(a[mid]);
    u->setc(build(l, mid - 1, a), 0);
    u->setc(build(mid + 1, r, a), 1);
    u->push_up();
    return u;
}
void debug(Node* r){
    if (r == null)return;
    printf("key:%I64d lson:%I64d rson:%I64d\n", r->key, r->ch[0]->key, r->ch[1]->key);
    debug(r->ch[0]);
    debug(r->ch[1]);
}

int n, q;
int a[N];
int main(){
    while (cin >> n >> q){
        for (int i = 1; i <= n; i++)rd(a[i]);
        init();
        Node* root = build(0, n + 1, a);
        Node* lnode,* rnode, *tmp;
        char str[2];
        while (q--){
            int l, r; ll val;
            scanf("%s", str); rd(l); rd(r);
            splay(root, get_left(root), null);
            lnode = get_kth(root, l);
            rnode = get_kth(root, r+2);
            splay(root, lnode, null);
    //      debug(root);
            splay(root, rnode, lnode);
    //      debug(root);
            if (str[0] == ‘Q‘)
            {
                pt(root->ch[1]->ch[0]->sum); putchar(‘\n‘);
            }
            else {
                rd(val);
                tmp = root->ch[1]->ch[0];
                tmp->sum += val * (r - l + 1);
                tmp->lazy += val;
                tmp->key += val;
                root->ch[1]->push_up();
                root->push_up();
        //      debug(root);
            }
        }
    }
    return 0;
}
/*
*/

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poj3468 A Simple Problem with Integers 指针版splay

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原文地址:http://blog.csdn.net/qq574857122/article/details/46743827

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