题目连接：

http://www.codeforces.com/contest/610/problem/D

Description

Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet and drew n black horizontal and vertical segments parallel to the coordinate axes. All segments have width equal to 1 square, that means every segment occupy some set of neighbouring squares situated in one row or one column.

Your task is to calculate the number of painted cells. If a cell was painted more than once, it should be calculated exactly once.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of segments drawn by Vika.

Each of the next n lines contains four integers x1, y1, x2 and y2 ( - 109 ≤ x1, y1, x2, y2 ≤ 109) — the coordinates of the endpoints of the segments drawn by Vika. It is guaranteed that all the segments are parallel to coordinate axes. Segments may touch, overlap and even completely coincide.

Output

Print the number of cells painted by Vika. If a cell was painted more than once, it should be calculated exactly once in the answer.

Sample Input

3

0 1 2 1

1 4 1 2

0 3 2 3

Sample Output

8

Hint

题意

在平面上会画n条宽度为1的线

然后问你最后画出来的线的总面积是多少

题解：

把画出来的线当成矩形

然后就是扫描线的裸题了

线段树跑一波就吼了

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define MID(a,b) (a+((b-a)>>1))
typedef long long LL;
const int N=360005;struct Line
{int x,y1,y2,flag;Line(){}Line(int a,int b,int c,int d){ x=a;y1=b;y2=c;flag=d; }bool operator<(const Line&b)const{ return x<b.x; }
};
struct node
{int lft,rht;int len[12],flag;int mid(){return MID(lft,rht);}void init(){memset(len,0,sizeof(len));}
};int n,k;
vector<int> y;
vector<Line> line;
map<int,int> H;struct Segtree
{node tree[N*4];void calu(int ind){if(tree[ind].flag>=k){int tmp=tree[ind].len[0];tree[ind].init();tree[ind].len[k]=tree[ind].len[0]=tmp;}else if(tree[ind].flag>0){int sum=0,cov=tree[ind].flag;for(int i=1;i<=k;i++) tree[ind].len[i]=0;tree[ind].len[cov]=tree[ind].len[0];if(tree[ind].lft+1==tree[ind].rht) return;for(int i=1;i<=k;i++){if(i+cov>=k) tree[ind].len[k]+=tree[LL(ind)].len[i]+tree[RR(ind)].len[i];else tree[ind].len[i+cov]=tree[LL(ind)].len[i]+tree[RR(ind)].len[i];}for(int i=cov+1;i<=k;i++) sum+=tree[ind].len[i];tree[ind].len[cov]-=sum;}else{for(int i=1;i<=k;i++) tree[ind].len[i]=0;if(tree[ind].lft+1==tree[ind].rht) return;for(int i=1;i<=k;i++)tree[ind].len[i]=tree[LL(ind)].len[i]+tree[RR(ind)].len[i];}}void build(int lft,int rht,int ind){tree[ind].lft=lft;  tree[ind].rht=rht;tree[ind].init();   tree[ind].flag=0;tree[ind].len[0]=y[rht]-y[lft];if(lft+1!=rht){int mid=tree[ind].mid();build(lft,mid,LL(ind));build(mid,rht,RR(ind));}}void updata(int st,int ed,int ind,int valu){int lft=tree[ind].lft,rht=tree[ind].rht;if(st<=lft&&rht<=ed) tree[ind].flag+=valu;else{int mid=tree[ind].mid();if(st<mid) updata(st,ed,LL(ind),valu);if(ed>mid) updata(st,ed,RR(ind),valu);}calu(ind);}
}seg;
int main()
{scanf("%d",&n);k=1;for(int i=0;i<n;i++){int x1,y1,x2,y2;scanf("%d%d%d%d",&x1,&y1,&x2,&y2);if(x1>x2)swap(x1,x2);if(y1>y2)swap(y1,y2);x2++;y2++;line.push_back(Line(x1,y1,y2,1));line.push_back(Line(x2,y1,y2,-1));y.push_back(y1); y.push_back(y2);}sort(line.begin(),line.end());sort(y.begin(),y.end());y.erase(unique(y.begin(),y.end()),y.end());for(int i=0;i<(int)y.size();i++) H[y[i]]=i;seg.build(0,(int)y.size()-1,1);LL res=0;for(int i=0;i<(int)line.size();i++){if(i!=0) res+=(LL)(line[i].x-line[i-1].x)*seg.tree[1].len[k];seg.updata(H[line[i].y1],H[line[i].y2],1,line[i].flag);}cout<<res<<endl;return 0;
}