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 xxyy ,( 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
3 1 4 4 1 3 3
样例输出
10
题目来源
题意：有n次海浪，每次海浪会产生两段轨迹(0,y)到(x,y)和(x,0)到(x,y)，后面的海浪会将前面的在(0,0)到(x,y)之间区域的海浪轨迹覆盖掉，问最后剩余的轨迹长度
分析：每段轨迹有效的贡献是在此轨迹被接下来所有轨迹覆盖一次后剩余的轨迹长度
所以我们考虑直接从后面往前面枚举
最后的轨迹长度肯定是可以直接加上的，没有其余的海浪能覆盖掉他，然后将其的轨迹点放进集合
接下来遍历到的点看集合只要看集合里比他小的最大的数就是能覆盖掉他多少长度，减去这个长度就行
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5+10;
const ll mod = 2e9+7;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll f( vector<ll> e ) {
ll sz = e.size(), ans = 0;
set<ll> s;
for( ll i = sz-1; i >= 0; i -- ) {
set<ll>::iterator it = s.lower_bound(e[i]);
if( it == s.begin() ) {
ans += e[i];
} else {
it --;
ans += e[i] - *it;
}
s.insert(e[i]);
}
return ans;
}
int main() {
ll x, y, n;
vector<ll> e1, e2;
scanf("%lld",&n);
while( n -- ) {
scanf("%lld%lld",&x,&y);
e1.push_back(x), e2.push_back(y);
}
printf("%lld\n",f(e1)+f(e2));
return 0;
}