Codeforces Round #337 (Div. 2) D. Vika and Segments 线段树 矩阵面积并

D. Vika and Segments

 

 

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

 

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：

　　给你n天平行x，y轴的线段

　　问你遍历的点有多少个

题解：

　　将线段 扩展成一个长度为x * 1 的矩阵

　　做一遍线段树扫描线求矩阵面积并

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")

const int N = 5e5+10, M = 5e5+11, inf = 2e9, mod = 1e9+7;
const double Pi = acos(-1.0);
typedef long long LL;
#define ls k<<1
#define rs ls | 1

int san[N], sum[N], vis[N], n, cnt = 0;
struct edge{
    int l,r,x,in;
    edge(int l = 0, int r = 0, int x = 0, int in = 0) : l(l), r(r), x(x), in(in) {}
    bool operator < (const edge &b) const {
        return x < b.x || x == b.x && in > b.in;
    }
}e[N];
int Hash(int x) {return lower_bound(san+1,san+cnt+1,x) - san;}
void push_up(int k,int ll,int rr) {
    if(vis[k]) sum[k] = san[rr + 1] - san[ll];
    else if(ll == rr) sum[k] = 0;
    else sum[k] = sum[ls] + sum[rs];
}
void update(int l,int r,int c,int ll,int rr,int k) {
    if(ll == l && rr == r) {
        vis[k] += c;
        push_up(k,ll,rr);
        return ;
    }
    int mid = (ll + rr) >> 1;
    if(r <= mid) update(l,r,c,ll,mid,ls);
    else if(l > mid) update(l,r,c,mid+1,rr,rs);
    else update(l,mid,c,ll,mid,ls), update(mid+1,r,c,mid+1,rr,rs);
    push_up(k,ll,rr);
}
int main() {
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i) {
                int x,y,xx,yy;
                scanf("%d%d%d%d",&x,&y,&xx,&yy);
                if(x > xx) swap(x,xx);
                if(y > yy) swap(y,yy);
                xx++, yy++;
                san[++cnt] = y;
                san[++cnt] = yy;
                e[i] = edge(y,yy,x,1);
                e[i+n] = edge(y,yy,xx,-1);
        }

        sort(san+1,san+cnt+1);
        cnt = unique(san + 1, san + cnt + 1) - san - 1;

        int m = n << 1;
        sort(e+1,e+m+1);

        LL ans = 0;
        for(int i = 1; i <= m; ++i) {
            int l = Hash(e[i].l);
            int r = Hash(e[i].r) - 1;
            if(l <= r) update(l,r,e[i].in,1,m,1);
            ans += 1LL * sum[1] * (e[i+1].x - e[i].x);
        }
        cout<<ans<<endl;
}