bzoj1056

treap+stl

这道题终于A了

排名是按照分数为第一关键字,插入时间为第二关键字,然后就是treap基本操作了

#include<bits/stdc++.h>
using namespace std;
const int N = 250010;
int n, Tim, root, score, rank;
map<string, pair<int, int> > info;
string ans[N];
char c[13];
struct Treap {
    int seed, cnt;
    pair<int, int> key[N];
    int size[N], child[N][2], p[N]; 
    int rand()
    {
        seed = abs(seed * 19992147 + 123456);
        return seed;
    }
    void update(int x)
    {
        size[x] = size[child[x][0]] + size[child[x][1]] + 1;
    }
    void rotate(int &x, int t)
    {
        int y = child[x][t];
        child[x][t] = child[y][t ^ 1];
        child[y][t ^ 1] = x;
        update(x);
        update(y);
        x = y;
    }
    void insert(int &x, pair<int, int> o)
    {
        if(x == 0)
        {
            p[x = ++cnt] = rand();
            key[x] = o;
            size[x] = 1;
        }
        else
        {
            int t = o > key[x];
            insert(child[x][t], o);
            if(p[child[x][t]] > p[x]) rotate(x, t);
        }
        update(x);
    }
    bool find(int x, pair<int, int> o)
    {
        if(x == 0) return false;
        if(key[x] == o) return true;
        return find(child[x][o > key[x]], o);
    }
    void erase(int &x, pair<int, int> o)
    {
        if(o == key[x]) 
        {
            if(child[x][0] == 0 && child[x][1] == 0)
            {
                x = 0;
                return;
            }
            int t = p[child[x][0]] < p[child[x][1]];
            rotate(x, t);
            erase(child[x][t ^ 1], o);
        }
        else erase(child[x][o > key[x]], o);
        update(x);
    }
    int query_name(int x, int rank)
    {
        if(size[child[x][1]] + 1 == rank) return -key[x].second;
        if(rank <= size[child[x][1]]) return query_name(child[x][1], rank);
        return query_name(child[x][0], rank - size[child[x][1]] - 1);        
    }
    int query_rank(int x, pair<int, int> o)
    {
        if(x == 0) return 0;
        if(o == key[x]) return size[child[x][1]] + 1;
        else if(o > key[x]) return query_rank(child[x][1], o);
        else return query_rank(child[x][0], o) + size[child[x][1]] + 1;
    }
} treap;
int main()
{
    scanf("%d", &n);
    while(n--)
    {
        pair<int, int> o;
        string s = "";
        scanf("%s", c);
        for(int i = 0; i < strlen(c); ++i) s = s + c[i];
        if(s[0] == '+')
        {
            scanf("%d", &score);
            s = s.substr(1, s.length() - 1);
            if(info.find(s) != info.end()) treap.erase(root, info[s]);
            info[s] = make_pair(score, -(++Tim));
            ans[Tim] = s;
            treap.insert(root, info[s]);
        }
        if(s[0] == '?') 
        {
            if(isdigit(s[1]))
            {
                rank = 0;
                for(int i = 1; i < s.length(); ++i) rank = rank * 10 + s[i] - '0';
                int lim = min(treap.size[root], rank + 9);
                for(int i = rank; i <= lim; ++i) 
                {
                    int id = treap.query_name(root, i), pos = 0;
                    while(isupper(ans[id][pos]))
                    {
                        putchar(ans[id][pos]);
                        ++pos;
                    }
                    if(i != lim) putchar(' ');
                }
                putchar('\n');
            }
            else
            {
                string name = s.substr(1, s.length() - 1);
                printf("%d\n", treap.query_rank(root, info[name]));
            }
        }        
    }
    return 0;
}
View Code

 

posted @ 2017-08-07 09:45  19992147  阅读(...)  评论(...编辑  收藏