Educational Codeforces Round 46 (Rated for Div. 2) C. Covered Points Count
Bryce1010模板
http://codeforces.com/problemset/problem/1000/C
题意:问你从[l,r]区间的被多少条线覆盖,列出所有答案。
思路:类似括号匹配的做法
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll MAXN=2e5+10;
struct Node
{
ll num;
ll dir;//0左1右
//ll cnt;
}node[MAXN*2];
ll cnt[MAXN*2];
bool cmp(Node a,Node b)
{
if(a.num!=b.num)return a.num<b.num;
else return a.dir<b.dir;
}
int main()
{
ll n;
cin>>n;
for(ll i=0;i<2*n;i++)
{
if(i&1)
{
cin>>node[i].num;
node[i].dir=1;
}
else
{
cin>>node[i].num;
node[i].dir=0;
}
}
sort(node,node+2*n,cmp);
// for(ll i=0;i<2*n;i++)
// cout<<node[i].num<<" "<<node[i].dir<<endl;;
ll tot=0;
for(ll i=0;i<2*n;i++)
{
if(node[i].dir==0)//左
{
if(node[i-1].dir==0)
cnt[tot]+=node[i].num-node[i-1].num;
else if(node[i-1].dir==1)
cnt[tot]+=node[i].num-node[i-1].num-1;
tot++;
}
else//右
{
if(node[i-1].dir==1)
cnt[tot]+=node[i].num-node[i-1].num;
else
cnt[tot]+=node[i].num-node[i-1].num+1;
tot--;
}
}
cout<<cnt[1];
for(ll i=2;i<=n;i++)
cout<<" "<<cnt[i];
cout<<endl;
return 0;
}
Educational Codeforces Round 46 (Rated for Div. 2) C. Covered Points Count的更多相关文章
- Educational Codeforces Round 50 (Rated for Div. 2) E. Covered Points
注释上都有解析了,就不写了吧,去重的问题就用set解决,并且呢第i个线段最多和其他线段产生i-1个交点,n^2logn. #include <cmath> #include <cst ...
- Educational Codeforces Round 46 (Rated for Div. 2) E. We Need More Bosses
Bryce1010模板 http://codeforces.com/contest/1000/problem/E 题意: 给一个无向图,求图的最长直径. 思路:对无向图缩点以后,求图的最长直径 #in ...
- Educational Codeforces Round 46 (Rated for Div. 2) B. Light It Up
Bryce1010模板 http://codeforces.com/problemset/problem/1000/B 思路:先用两个数组sumon[]和sumoff[]将亮着的灯和灭的灯累计一下. ...
- Educational Codeforces Round 46 (Rated for Div. 2) A. Codehorses T-shirts
Bryce1010模板 http://codeforces.com/problemset/problem/1000/A 题意: 问你将一种类型的衣服转换成另一种的最小次数. #include<b ...
- Educational Codeforces Round 46 (Rated for Div. 2)
A - Codehorses T-shirts 思路:有相同抵消,没有相同的对答案+1 #include<bits/stdc++.h> #define LL long long #defi ...
- Educational Codeforces Round 46 (Rated for Div. 2) D. Yet Another Problem On a Subsequence
这个题是dp, dp[i]代表以i开始的符合要求的字符串数 j是我们列举出的i之后一个字符串的开始地址,这里的C是组合数 dp[i] += C(j - i - 1, A[i]] )* dp[j]; # ...
- Educational Codeforces Round 46 (Rated for Div. 2) D
dp[i]表示一定包含第I个点的好的子序列个数,那么最终答案就是求dp[0] + dp[1] + .... + dp[n-1] 最终的子序列被分成了很多块,因此很明显我们枚举第一块,第一块和剩下的再去 ...
- Educational Codeforces Round 72 (Rated for Div. 2)-D. Coloring Edges-拓扑排序
Educational Codeforces Round 72 (Rated for Div. 2)-D. Coloring Edges-拓扑排序 [Problem Description] 给你 ...
- Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship
Problem Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...
随机推荐
- String、StringBilder和StringBuffer之间的区别
1.三者在执行速度方面的比较:StringBuilder > StringBuffer > String 2.String <(StringBuffer,StringBuild ...
- 笔记整理--LibCurl开发
LibCurl开发_未了的雨_百度空间 - Google Chrome (2013/7/26 21:11:15) LibCurl开发 一:LibCurl 编程流程1.调用curl_global_ini ...
- Apach Web Server区别于其他应用服务器的主要特点是什么?
Web服务器一般指的是处理静态请求或转发http请求的服务器,而应用服务器一般是用来处理动态请求的服务器.
- linux下如何用php读取word
在实际的工作中遇到到要导入word格式的文件,经过努力,终于成功了. 在linux上用PHP读取WORD文档,其实是使用了 antiword程序把word文档转化为txt文档. 再使用php执行系统命 ...
- XMU C语言程序设计实践(5)
• 使用动态链表完成一个简单的商品库存信息管理系统. • 商品信息包括如下字段:商品号.商品名称.商品库存 • 函数 create:接收用户输入的商品号和商品名称的 ...
- INFO: Ignoring response <403 https://movie.douban.com/top250>: HTTP status code is not handled or not allowed
爬取豆瓣电影top250,出现以下报错: 2018-08-11 22:02:16 [scrapy.core.engine] INFO: Spider opened 2018-08-11 22:02:1 ...
- 解决VMware安装Ubuntu的过程中窗口过小无法看到【下一步】按钮的问题
只要按住ALT键向上拖动窗口
- Simple JavaScript Inheritance
1. [代码]Simple JavaScript Inheritance (function(){ var initializing = false, fnTest = /xyz/.test ...
- zoj 3204 Connect them(最小生成树)
题意:裸最小生成树,主要是要按照字典序. 思路:模板 prim: #include<iostream> #include<stdio.h> #include<string ...
- Intel® Media SDK(一)
A cross-platform API for developing media applications on Windows* Fast video playback, encode, proc ...