海浪覆盖问题，但不存在一个海浪被另一个海浪完全覆盖的问题，所以我们可以从最后一个海浪求，若最后一个海浪的宽度为x,然后以他为起点，可以用一个set来存储，宽度比set的首元素小的直接加上，比他大的加上他们的差值，没比较一个结果就加上set（关键就是没有完全被覆盖的）
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
代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<set>
#include<string>
#include<vector>
using namespace std;
long long gao(vector<int> vec) {
int sz = vec.size();
set<int>st;
long long ans = 0;
for (int i = sz - 1; i >= 0; i--) {
set<int>::iterator it = st.lower_bound(vec[i]);
if (it == st.begin()) {
ans += vec[i];
}
else {
it--;
ans += vec[i] - *it;
}
st.insert(vec[i]);
}
return ans;
}
int main() {
int n;
while (scanf("%d", &n) == 1) {
vector<int>vec1, vec2;
int x, y;
while (n--) {
scanf("%d%d", &x, &y);
vec1.push_back(x);
vec2.push_back(y);
}
cout << gao(vec1) + gao(vec2) << endl;
}
}