vijosp1507郁闷的出纳员

一道平衡树实现的名次树,用了treap实现。

1.rand()产生的数可能比INF大,很可能改变树的结构。

2.删除时先递归,然后再删除自己,实现就变得简单多了。

3.因为很多情况树会是空的,所以设了一个root虚拟节点。设指针时一定要new一个出来。

就这样,其实水题一道。

#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
const int maxn = 100000 + 10;
const int INF = 1000000000 + 7;

int n,minv,p = 0,ans = 0,sum;

struct Node {
    int v,r,s;
    Node* ch[2];
    
    int cmp(int x) const {
        if(v == x) return -1;
        return v < x ? 0 : 1;
    }
    
    void maintain() {
        s = 1;
        if(ch[0] != NULL) s += ch[0]->s;
        if(ch[1] != NULL) s += ch[1]->s;
    }
    
} a[maxn];

Node *root = new Node();

void 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;
}

void insert(Node* &o,int x) {
    if(o == NULL) {
        o = &a[p++];
        o->v = x;
        o->ch[0] = o->ch[1] = NULL;
        o->r = rand()%INF;
        o->s = 1;
    }
    else {
        int d = o->v > x ? 0:1;
        insert(o->ch[d],x);
        if(o->ch[d]->r > o->r) rotate(o,d^1);
        o->maintain();
    }
}

void remove(Node* &o,int x) {
    if(o == NULL) return;
    if(o->v >= x) {
        remove(o->ch[0],x);
        o->maintain();
    }
    else {
        remove(o->ch[0],x);
        remove(o->ch[1],x);
        if(o->ch[1] == NULL) {
            o = NULL;
            ans++;
        }
        else {
            rotate(o,0);
            remove(o->ch[0],x);
            o->maintain();
        }
    }
}

int find(Node* o,int k) {
    if(o == root) {return find(o->ch[0],k);}
    else {
        if(o == NULL) return (-1);
        int m;
        if(o->ch[1] == NULL) m = 0;
        else m = o->ch[1]->s;
        if(k == m+1) return o->v;
        else if(k < m+1) return find(o->ch[1],k);
        else return find(o->ch[0],k-m-1);
    }
}

char s[10];

int main() {
    scanf("%d%d",&n,&minv);
    root->v = root->r = INF;
    root->s = 1;
    while(n--) {
        int v;
        scanf("%s%d",s,&v);
        if(s[0] == 'I') if(v >= minv) {insert(root,v-sum);};
        if(s[0] == 'A') sum += v;
        if(s[0] == 'S') {
            sum -= v;
            remove(root,minv-sum);
        }
        if(s[0] == 'F') {
            if(root->s-1 < v) printf("-1
");
            else printf("%d
",find(root,v)+sum);
        }
    }
    printf("%d
",ans);
    return 0;
}
原文地址:https://www.cnblogs.com/invoid/p/5325349.html