codeforces 610D D. Vika and Segments(离散化+线段树+扫描线算法)

题目链接：

D. Vika and Segments

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

3
0 1 2 1
1 4 1 2
0 3 2 3

output

8

input

4
-2 -1 2 -1
2 1 -2 1
-1 -2 -1 2
1 2 1 -2

output

16

Note

In the first sample Vika will paint squares (0, 1), (1, 1), (2, 1), (1, 2), (1, 3), (1, 4), (0, 3) and (2, 3).

题意：

给了这么多线段,问它们一共包含了多少个点;

思路：

把线段变成宽为1的矩形,然后用扫描线算法求面积;

AC代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+4;
int n,x1,x2,y3,y2,rec[2*N],num;
struct no
{int l,r,h,flag;
};
no line[8*N];
struct nod
{int l,r,cover;ll sum;
};
nod tree[8*N];
int cmp(no x,no y)
{return x.h<y.h;
}
void build(int node,int L,int R)
{tree[node].l=L,tree[node].r=R;tree[node].cover=tree[node].sum=0;if(L>=R)return ;int mid=(L+R)>>1;build(2*node,L,mid);build(2*node+1,mid+1,R);
}
void Pushup(int node)
{if(tree[node].cover){tree[node].sum=rec[tree[node].r+1]-rec[tree[node].l];}else{if(tree[node].l==tree[node].r)tree[node].sum=0;else tree[node].sum=tree[2*node].sum+tree[2*node+1].sum;}
}
void update(int node,int L,int R,int x)
{if(L<=tree[node].l&&R>=tree[node].r){tree[node].cover+=x;Pushup(node);return ;}int mid=(tree[node].l+tree[node].r)>>1;if(L>mid) update(2*node+1,L,R,x);else if(R<=mid)update(2*node,L,R,x);else{update(2*node,L,mid,x);update(2*node+1,mid+1,R,x);}Pushup(node);
}
int bi(int x)
{int L=1,R=num-1,mid;while(L<=R){mid=(L+R)>>1;if(rec[mid]==x)return mid;else if(rec[mid]>x)R=mid-1;else L=mid+1;}return -1;
}
int main()
{scanf("%d",&n);int cnt=1;for(int i=1;i<=n;i++){scanf("%d%d%d%d",&x1,&y3,&x2,&y2);if(x1>x2)swap(x1,x2);if(y3>y2)swap(y3,y2);rec[cnt] = line[cnt].l = x1;line[cnt].r = x2+1;line[cnt].h = y3;line[cnt++].flag = 1;line[cnt].l = x1;rec[cnt] = line[cnt].r = x2+1;line[cnt].h = y2+1;line[cnt++].flag = -1;}sort(line+1,line+cnt,cmp);sort(rec+1,rec+cnt);num = 2;for(int i = 2;i < cnt;i++){if(rec[i]!=rec[i-1])rec[num++]=rec[i];}build(1,1,num-1);ll ans=0;for(int i = 1;i < cnt-1;i++){int fx = bi(line[i].l);int fy = bi(line[i].r)-1;if(fx <= fy){update(1,fx,fy,line[i].flag);}ans+=tree[1].sum*(ll)(line[i+1].h-line[i].h);}cout<<ans<<"\n";return 0;
}