ACM-ICPC 2018 徐州赛区网络预赛 G. Trace

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 0 , 0 ), ( x , 0 ), ( 0 , y ), ( x ,y ). Every time the wave will wash out the trace of former wave in its range and remain its own trace of ( x , 0 ) -> (x ,y ) and (0 , y ) -> ( x , y ). Now the toad on the coast wants to know the total length of trace on the coast after n waves. It's guaranteed that a wave will not cover the other completely.

Input

The first line is the number of waves n(n≤50000).

The next nn lines，each contains two numbers x y ,( 0<x , y≤10000000 )，the ii-th line means the i-th second there comes a wave of ( xx , yy ), it's guaranteed that when 1≤i , j≤n ，xi​≤xj​ and jyi​≤yj​ don't set up at the same time.

Output

An Integer stands for the answer.

Hint：

As for the sample input, the answer is 3+3+1+1+1+1=103+3+1+1+1+1=10

样例输入复制

3
1 4
4 1
3 3

样例输出复制

10

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

 

   //分成横纵坐标来考虑
//如 ：纵坐标，因为题目已明确会覆盖，但不会完全覆盖
//那么，(会影响我的)比我高的一定在我左边(不然完全覆盖)，比我矮的
//一定在我右边，(不然就完全不覆盖了)。
#include<bits/stdc++.h>
using  namespace  std;
#define  ll long long 
#define N 50009
int n;
ll x[N],y[N];
ll solve(ll a[])
{
    set<ll>se;
    set<ll>::iterator it;
    ll ans=0;
    for(int i=n;i>=1;i--)
    {
        it=se.lower_bound(a[i]);
        if(it==se.begin())  {//不存在，或者都比我大
            ans+=a[i];
        }
        else{
            it--;//不可能有2 2 1 3
            ans+=a[i]-(*it);
        }
        se.insert(a[i]);
    }
    return  ans;
}
int  main()
{
    while(~scanf("%d",&n))
    {    
        for(int i=1;i<=n;i++)
        {
            scanf("%lld%lld",&x[i],&y[i]);
        }
        printf("%lld\n",solve(x)+solve(y));
    }
    return 0;
}