HDU 1828 Picture（线段树扫描线求周长）

Problem Description

A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of all rectangles is called the perimeter.

Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1.

The corresponding boundary is the whole set of line segments drawn in Figure 2.

The vertices of all rectangles have integer coordinates.

Input

Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate.

0 <= number of rectangles < 5000
All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.

Please process to the end of file.

Output

Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles.

Sample Input

Sample Output

#include <stdio.h>

#include <bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define CLR(arr,val) memset(arr,val,sizeof(arr))

#define LC(x) (x<<1)

#define RC(x) ((x<<1)+1)

#define MID(x,y) ((x+y)>>1)

typedef pair<int,int> pii;

typedef long long LL;

const double PI=acos(-1.0);

const int N=2e4+7;

struct seg

{

    int l,mid,r;

    int len,cnt;

};

struct Line

{

    int l,r,h,flag;

    inline bool operator<(const Line &t)const

    {

        if(h==t.h)//若高度相等，则底边优先

            return flag>t.flag;

        return h<t.h;

    }

};

seg T[N<<2];

Line xline[N],yline[N];

inline void pushup(int k)

{

    if(T[k].cnt)

        T[k].len=T[k].r-T[k].l+1;

    else

    {

        if(T[k].l==T[k].r)

            T[k].len=0;

        else

            T[k].len=T[LC(k)].len+T[RC(k)].len;

    }

}

void build(int k,int l,int r)

{

    T[k].l=l;

    T[k].r=r;

    T[k].mid=MID(l,r);

    T[k].cnt=0;

    T[k].len=0;

    if(l==r)

        return ;

    build(LC(k),l,T[k].mid);

    build(RC(k),T[k].mid+1,r);

}

void update(int k,int l,int r,int flag)

{

    if(l<=T[k].l&&T[k].r<=r)

    {

        T[k].cnt+=flag;

        pushup(k);

    }

    else

    {

        if(r<=T[k].mid)

            update(LC(k),l,r,flag);

        else if(l>T[k].mid)

            update(RC(k),l,r,flag);

        else

            update(LC(k),l,T[k].mid,flag),update(RC(k),T[k].mid+1,r,flag);

        pushup(k);

    }

}

int main(void)

{

    int n,i;

    int xa,xb,ya,yb;

    int ans,last;

    while (~scanf("%d",&n))

    {

        int cnt=0;

        int xR=-INF;

        int yR=-INF;

        int xL=INF;

        int yL=INF;

        for (i=0; i<n; ++i)

        {

            scanf("%d%d%d%d",&xa,&ya,&xb,&yb);

            xline[cnt]=(Line){xa,xb,ya,1};

            yline[cnt++]=(Line){ya,yb,xa,1};//

            xline[cnt]=(Line){xa,xb,yb,-1};

            yline[cnt++]=(Line){ya,yb,xb,-1};//

            if(xa>xR) xR=xa;

            if(xa<xL) xL=xa;

            if(xb>xR) xR=xb;

            if(xb<xL) xL=xb;

            if(ya>yR) yR=ya;

            if(ya<yL) yL=ya;

            if(yb>yR) yR=yb;

            if(yb<yL) yL=yb;

        }

        ans=0;

        sort(xline,xline+cnt);

        sort(yline,yline+cnt);

        last=0;

        build(1,xL,xR);

        for (i=0; i<cnt; ++i)

        {

            update(1,xline[i].l,xline[i].r-1,xline[i].flag);

            ans=ans+abs(T[1].len-last);

            last=T[1].len;

        }

        last=0;

        build(1,yL,yR);

        for (i=0; i<cnt; ++i)

        {

            update(1,yline[i].l,yline[i].r-1,yline[i].flag);

            ans=ans+abs(T[1].len-last);

            last=T[1].len;

        }

        printf("%d\n",ans);

    }

    return 0;

}