UVA-11020(BST)

题意：

给n个点,一个点(x,y)有优势时满足不存在点(fx,fy)使得fx<x,fy<=y或fx<=x,fy<y;问当前有多少个有优势点;

思路：

学习BST的入门题,代码是白书上的;

AC代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <bits/stdc++.h>

#include <stack>

#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)

#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {

    char CH; bool F=false;

    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());

    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());

    F && (num=-num);

}

int stk[70], tp;

template<class T> inline void print(T p) {

    if(!p) { puts("0"); return; }

    while(p) stk[++ tp] = p%10, p/=10;

    while(tp) putchar(stk[tp--] + '0');

    putchar('\n');

}

const LL mod=1e9+7;

const double PI=acos(-1.0);

const int inf=1e9+10;

const int N=1e5+10;

const int maxn=1e3+20;

const double eps=1e-12;

struct PO

{

    int x,y;

    bool operator< (const PO& rhs) const

    {

        return x<rhs.x||(x==rhs.x&&y<rhs.y);

    }

};

multiset<PO>s;

multiset<PO>::iterator it;

int main()

{

    int t,Case=0;

    read(t);

    while(t--)

    {

        s.clear();

        if(Case)printf("\n");

        printf("Case #%d:\n",++Case);

        int n,x,y;

        read(n);

        For(i,1,n)

        {

            read(x);read(y);

            PO temp=(PO){x,y};

            it=s.lower_bound(temp);

            if(it == s.begin()||(--it)->y > y)

            {

                s.insert(temp);

                it=s.upper_bound(temp);

                while(it!=s.end()&&it->y >= y)s.erase(it++);

            }

            printf("%d\n",s.size());

        }

    }

    return 0;

}