#include <bits/stdc++.h>
using namespace std; #define pb push_back
#define lb lower_bound
#define ull unsigned ll
#define gcd(a,b) __gcd(a,b)
#define pii pair<int,int>
#define all(x) x.begin(),x.end()
#define ll long long
#define mp make_pair
//#define pi acos(-1) #define mod 1000000007
#define inf (1LL<<31)-1 map<int,vector<pii> > X,Y;
vector<pair<int,pii> > V; const int mx = 200005;
int tree[mx], pos[mx],N; int getPos(int n)
{
return (int) (lb(pos,pos+N,n)-pos) + 1;
} int update(int idx,int val)
{
while(idx<=N)
{
tree[idx] += val;
idx += (idx&-idx);
}
return 0;
} int sum(int idx)
{
int res = 0;
while(idx>0)
{
res += tree[idx];
idx -= (idx&-idx);
}
return res;
} int main()
{
int i,j,k,n;
scanf("%d",&n);
int x1,y1,x2,y2;
for(i=0;i<n;i++)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
if(x1>x2) swap(x1,x2);
if(y1>y2) swap(y1,y2);
if(x1==x2) Y[x1].pb(mp(y1,y2));
else X[y1].pb(mp(x1,x2));
pos[N++] = y1;
pos[N++] = y2;
}
sort(pos,pos+N);
N = (int) (unique(pos,pos+N)-pos); map<int,vector<pii> > :: iterator it; ll ans = 0;
for(it=X.begin();it!=X.end();it++)
{
vector<pii>& vt = it -> second;
sort(all(vt));
int y = (it -> first);
int x1 = vt[0].first;
int x2 = vt[0].second;
int sz = vt.size();
for(i=1;i<sz;i++)
{
if(x2<vt[i].first)
{
ans += (x2-x1+1);
V.pb(mp(x1,mp(-1,y)));
V.pb(mp(x2+1,mp(1,y)));
x1 = vt[i].first;
x2 = vt[i].second;
}
else x2 = max(x2,vt[i].second);
}
ans += (x2-x1+1);
V.pb(mp(x1,mp(-1,y)));
V.pb(mp(x2+1,mp(1,y)));
} int m = V.size(), p = 0;
sort(all(V));
for(it=Y.begin();it!=Y.end();it++)
{
vector<pii>& vt = it -> second;
sort(all(vt));
int x = (it -> first); while(p<m&&x>=V[p].first)
{
int t1 = V[p].second.first;
int t2 = V[p].second.second;
t2 = getPos(t2);
update(t2,-t1);
p++;
} int y1 = vt[0].first;
int y2 = vt[0].second;
int sz = vt.size();
for(i=1;i<sz;i++)
{
if(y2<vt[i].first)
{
ans += (y2-y1+1);
y2 = getPos(y2);
y1 = getPos(y1);
ans -= sum(y2) - sum(y1-1);
y2 = vt[i].second;
y1 = vt[i].first;
}
else y2 = max(y2,vt[i].second);
}
ans += (y2-y1+1);
y2 = getPos(y2);
y1 = getPos(y1);
ans -= sum(y2) - sum(y1-1);
}
cout << ans << endl;
}

  

codeforce div 377的更多相关文章

  1. CodeForce Div 2 C. Masha and two friends

    题目链接: http://codeforces.com/contest/1080/problem/C 思路:双向延长两个矩形方块的4边,会形成一个被分割为9块的更大立方体. 计算所有的9个方框.方框中 ...

  2. Codeforces Round #377 (Div. 2) D. Exams

    Codeforces Round #377 (Div. 2) D. Exams    题意:给你n个考试科目编号1~n以及他们所需要的复习时间ai;(复习时间不一定要连续的,可以分开,只要复习够ai天 ...

  3. Codeforces Round #377 (Div. 2)

    #include <iostream> #include <stdio.h> #include <string.h> using namespace std; in ...

  4. codeforce AIM tech Round 4 div 2 B rectangles

    2017-08-25 15:32:14 writer:pprp 题目: B. Rectangles time limit per test 1 second memory limit per test ...

  5. Codeforces Round #377 (Div. 2) B. Cormen — The Best Friend Of a Man(贪心)

     传送门 Description Recently a dog was bought for Polycarp. The dog's name is Cormen. Now Polycarp has ...

  6. Codeforces Round #377 (Div. 2) D. Exams(二分答案)

    D. Exams Problem Description: Vasiliy has an exam period which will continue for n days. He has to p ...

  7. Codeforces Round #377 (Div. 2)D(二分)

    题目链接:http://codeforces.com/contest/732/problem/D 题意: 在m天中要考k个课程, 数组a中有m个元素,表示第a[i]表示第i天可以进行哪门考试,若a[i ...

  8. Codeforce#331 (Div. 2) A. Wilbur and Swimming Pool(谨以此题来纪念我的愚蠢)

    C time limit per test 1 second memory limit per test 256 megabytes input standard input output stand ...

  9. Codeforces Round #377 (Div. 2) C. Sanatorium 水题

    C. Sanatorium time limit per test 1 second memory limit per test 256 megabytes input standard input ...

随机推荐

  1. LintCode Singleton

    Singleton 3 大要素: 1.有private static的句柄(成员变量即field) 2. constructor 必须为private 3.有public static的getInst ...

  2. 2015GitWebRTC编译实录14

    libvpx 尝试用脚本编译了下,发现有问题,就偃旗息鼓,改用他自己的configure了,在网上找了下,Git上有个现成的,直接用,更好些. https://github.com/brion/VPX ...

  3. excel导入导出

    using System; using System.Collections.Generic; using System.Linq; using System.Text; using FS.Exten ...

  4. ASP.NET 下拉列表绑定枚举类型值,不用再新建一个枚举表

    public partial class Default : System.Web.UI.Page { protected void Page_Load(object sender, EventArg ...

  5. HTML适应手机浏览器宽度

    <meta name="viewport" content="width=device-width, initial-scale=1.0, minimum-scal ...

  6. try-catch(C# 参考)

    https://msdn.microsoft.com/zh-cn/library/0yd65esw.aspx Try-catch 语句包含一个后接一个或多个 catch 子句的 try 块,这些子句指 ...

  7. python(23)re函数:compile、match、search、findall

    正则表达式是一个特殊的字符序列,它能帮助你方便的检查一个字符串是否与某种模式匹配. Python 自1.5版本起增加了re 模块,它提供 Perl 风格的正则表达式模式. re 模块使 Python ...

  8. SQL Error (1130): Host '192.168.1.126' is not allowed to connect to this MySQL server

    通过HeidiSQL连接MYSQL数据库报错: SQL Error (1130): Host '192.168.1.126' is not allowed to connect to this MyS ...

  9. flask 添加日志

    def add_error_handler(app): for exception in default_exceptions: app.register_error_handler(exceptio ...

  10. JNI环境搭建,CDT, cygwin,NDK

    1.为eclipse增加c和c++的开发插件 下载地址:http://www.eclipse.org/cdt/downloads.php 2,安装cygwin Ndk需要运行在linux环境下,cyg ...