2018徐州icpc网络赛B题

最新推荐文章于 2019-09-07 23:16:21 发布

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最新推荐文章于 2019-09-07 23:16:21 发布

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本文介绍了一种计算经过多次海浪冲刷后海岸线上痕迹总长度的方法。通过使用数据结构set来跟踪每次海浪冲击后的海岸线变化，并确保没有一次海浪会完全覆盖另一次海浪。算法最终返回海岸线痕迹的总长度。

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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

样例输出复制

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;
	}
}