Trace

最新推荐文章于 2025-02-24 23:05:30 发布

咕咕评测姬

最新推荐文章于 2025-02-24 23:05:30 发布

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分类专栏：多校

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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

样例输入复制

样例输出复制

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

#include<bits/stdc++.h>
using namespace std;
int n;
long long dfs(vector<long long>w) {
    long long sum=0;
    set<int>s;
    set<int>::iterator it;
    for(int i=n-1;i>=0;i--) {
        it=s.lower_bound(w[i]);//找到在它之前大于等于的，即在这之前是被覆盖的 
        if(it==s.begin()){//没有比它小的则不被覆盖直接加 
            sum += w[i];
        } 
		else{
            it--;
            sum+=w[i]-*it;
        }
        s.insert(w[i]);
    }
    return sum;
}
int main() {
	long long x,y;
    vector<long long> we,ww;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%ld %ld",&x,&y);
        we.push_back(x);//横坐标 
		ww.push_back(y);//纵坐标 
    }
    printf("%lld\n",dfs(we)+dfs(ww));
    return 0;
}