bzoj2658: [Zjoi2012]小蓝的好友(mrx)

太神辣 treap的随机键值竟然能派上用场。。

要用不旋转的treap来进行维护区间信息

 

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>

using namespace std;

template<typename Q> Q &read(Q &x) {
    static char c, f;
    for(f = 0; c = getchar(), !isdigit(c); ) if(c == '-') f = 1;
    for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    if(f) x = -x; return x;
}
template<typename Q> Q read() {
    static Q x; read(x); return x;
}

typedef long long LL;
const int N = 40000 + 10;

LL S(LL x) {
    return x * (x + 1) >> 1;
}

struct Node *null, *pis;
struct Node {
    int sz, h, tag;
    LL ans;
    Node *ch[2];
    
    Node() {}
    Node(int h) : h(h) {
        sz = 1, ans = tag = 0;
        ch[0] = ch[1] = null;
    }
    
    void add(int d) {
        if(this == null) return;
        h += d, tag += d;
    }
    
    void maintain() {
        sz = ch[0]->sz + ch[1]->sz + 1;
        ans = 0;
        for(int c = 0; c < 2; c++) {
            ans += ch[c]->ans + S(ch[c]->sz) * (ch[c]->h - h);
        }
    }
    
    void down() {
        ch[0]->add(tag);
        ch[1]->add(tag);
        tag = 0;
    }
    
    void *operator new(size_t) {
        return pis++;
    }
}pool[N];

Node *merge(Node *l, Node *r) {
    if(l == null) return r;
    if(r == null) return l;
    if(l->h < r->h) {
        l->down();
        l->ch[1] = merge(l->ch[1], r);
        return l->maintain(), l;
    }else {
        r->down();
        r->ch[0] = merge(l, r->ch[0]);
        return r->maintain(), r;
    }
}

typedef pair<Node *, Node *> pnn;
pnn split(Node *o, int k) {
    if(o == null) return pnn(null, null);
    pnn res; o->down();
    if(o->ch[0]->sz >= k) {
        res = split(o->ch[0], k);
        o->ch[0] = res.second;
        res.second = o;
    }else {
        res = split(o->ch[1], k - o->ch[0]->sz - 1);
        o->ch[1] = res.first;
        res.first = o;
    }
    return o->maintain(), res;
}

pair<int, int> p[100000 + 10];

int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    
    int r, c, n;
    scanf("%d%d%d", &r, &c, &n);
    for(int i = 0; i < n; i++) {
        scanf("%d%d", &p[i].first, &p[i].second);
    }
    sort(p, p + n);
    
    pis = pool, null = new Node(0);
    null->sz = 0;
    Node *root = null;
    for(int i = 1; i <= c; i++) {
        root = merge(root, new Node(0));
    }
    
    LL ans = S(r) * S(c);
    for(int i = 1, j = 0; i <= r; i++) {
        root->add(1);
        while(j < n && p[j].first == i) {
            int x = p[j++].second;
            pnn r1 = split(root, x - 1);
            pnn r2 = split(r1.second, 1);
            r2.first->h = 0;
            root = merge(merge(r1.first, r2.first), r2.second);
        }
        ans -= root->ans + S(root->sz) * root->h;
    }
    cout << ans << endl;
    
    return 0;
}

 

谁说一定要用fhq treap？

用普通的treap就好了 而且常数小！

注意建树最好$O(n)$建一下，每次insert有可能退化成$O(n^2)$的。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
 
using namespace std;
 
template<typename Q> Q &read(Q &x) {
    static char c, f;
    for(f = 0; c = getchar(), !isdigit(c); ) if(c == '-') f = 1;
    for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    if(f) x = -x; return x;
}
template<typename Q> Q read() {
    static Q x; read(x); return x;
}
 
typedef long long LL;
const int N = 40000 + 10;
 
LL S(LL x) {
    return x * (x + 1) >> 1;
}
 
struct Node *null, *pis, *bin[N];
int top;
struct Node {
    int sz, v, h, tag;
    LL ans;
    Node *ch[2];
     
    Node() {}
    Node(int v, int h) : v(v), h(h) {
        sz = 1, ans = tag = 0;
        ch[0] = ch[1] = null;
    }
     
    void add(int d) {
        if(this == null) return;
        h += d, tag += d;
    }
     
    void maintain() {
        sz = ch[0]->sz + ch[1]->sz + 1;
        ans = 0;
        for(int c = 0; c < 2; c++) {
            ans += ch[c]->ans + S(ch[c]->sz) * (ch[c]->h - h);
        }
    }
     
    void down() {
        ch[0]->add(tag);
        ch[1]->add(tag);
        tag = 0;
    }
     
    void *operator new(size_t) {
        return top ? bin[--top] : pis++;
    }
    
    void operator delete(void *p) {
        bin[top++] = (Node *) p;
    }
    
    int cmp(int x) const {
        if(x == v) return -1;
        return x < v ? 0 : 1;
    }
}pool[N];

void rotate(Node *&o, int d) {
    Node *t = o->ch[d];
    o->ch[d] = t->ch[d ^ 1];
    t->ch[d ^ 1] = o;
    o->maintain();
    (o = t)->maintain();
}

void modify(Node *&o, int x, int w) {
    o->down();
    int d = o->cmp(x);
    if(d == -1) return o->h = w, o->maintain(), void();
    modify(o->ch[d], x, w);
    if(o->ch[d]->h < o->h) rotate(o, d);
    else o->maintain();
}

pair<int, int> p[100000 + 10];

void build(Node *&o, int l, int r) {
    if(l > r) return;
    int mid = (l + r) >> 1;
    o = new Node(mid, 0);
    build(o->ch[0], l, mid - 1);
    build(o->ch[1], mid + 1, r);
    o->maintain();
}

int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
     
    int r, c, n;
    scanf("%d%d%d", &r, &c, &n);
    for(int i = 0; i < n; i++) {
        read(p[i].first), read(p[i].second);
    }
    sort(p, p + n);
     
    pis = pool, null = new Node(0, 0);
    null->sz = 0;
    Node *root;
    build(root, 1, c);
    
    LL ans = S(r) * S(c);
    for(int i = 1, j = 0; i <= r; i++) {
        root->add(1);
        while(j < n && p[j].first == i) {
            modify(root, p[j++].second, 0);
        }
        ans -= root->ans + S(root->sz) * root->h;
    }
    cout << ans << endl;
     
    return 0;
}