https://www.jisuanke.com/contest/1557?view=challenges
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 ( 00 , 00 ), ( xx , 00 ), ( 00 , yy ), ( xx , yy ). Every time the wave will wash out the trace of former wave in its range and remain its own trace of ( xx , 00 ) -> ( xx , yy ) and ( 00 , yy ) -> ( xx , yy ). 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 \le 50000)n(n≤50000).
The next nn lines，each contains two numbers xx yy ,( 0 < x0<x , y \le 10000000y≤10000000 )，the ii-th line means the ii-th second there comes a wave of ( xx , yy ), it's guaranteed that when 1 \le i1≤i , j \le nj≤n ，x_i \le x_jxi≤xj and y_i \le y_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
题意：后个海浪会覆盖前个海浪的一部分，所求轨迹长度就是矩形边缘线的长度（矩形左上角到右上角的长度+矩形右下角到右上角的长度）。
题解：对x坐标和y坐标分别计算。用vector分别记录x，y坐标，从最后一波海浪往前计算，如果在set中能找到第一个比这波海浪x值小的，那么ans就加上二者差值，否则ans就加上当前海浪的x值，y同样如此
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<vector>
using namespace std;
#define inf 0x3f3f3f
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a));
#define lowbit(x) x&-x;
typedef long long ll;
const int maxn = 1e5+5;
int n,q;
set<int>st;
vector<int>x,y;
int main(){
while(~scanf("%d",&n)){
st.clear();
x.clear(),y.clear();
int l,r;
for(int i = 1; i <= n; i++){
scanf("%d%d",&l,&r);
x.push_back(l);
y.push_back(r);
}
ll ans = 0;
for(int i = x.size()-1; i >= 0; i--){
auto it = st.lower_bound(x[i]);
if(it != st.begin()){ //找到第一个比x[i]大的数
it--; //减后就是第一个比x[i]小的数
ans += (x[i] - *it);
}else{
ans += x[i];
}
st.insert(x[i]);
}
st.clear();
for(int i = y.size()-1; i >= 0; i--){
auto it = st.lower_bound(y[i]);
if(it != st.begin()){
it--;
ans += (y[i] - *it);
}else{
ans += y[i];
}
st.insert(y[i]);
}
cout<<ans<<endl;
}