POJ训练计划1177_Picture(扫描线/线段树+离散)

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解题报告

题意：

求矩形周长和。

思路：

左扫上扫，扫过了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
struct Seg {
    int lx,rx,ly,ry,h,v;
    friend bool operator < (Seg a,Seg b)
    {
        return a.h<b.h;
    }
} seg1[11000],seg2[11000];
int _hx[21000],_hy[21000],sum[500000],lz[500000];
void push_up1(int rt,int l,int r)
{
    if(lz[rt]) {
        sum[rt]=_hx[r+1]-_hx[l];
    } else sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void update1(int rt,int l,int r,int ql,int qr,int v)
{
    if(ql>r||qr<l)return ;
    if(ql<=l&&r<=qr) {
        lz[rt]+=v;
        push_up1(rt,l,r);
        return ;
    }
    int mid=(l+r)>>1;
    update1(rt<<1,l,mid,ql,qr,v);
    update1(rt<<1|1,mid+1,r,ql,qr,v);
    push_up1(rt,l,r);
}
void push_up2(int rt,int l,int r)
{
    if(lz[rt]) {
        sum[rt]=_hy[r+1]-_hy[l];
    } else sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void update2(int rt,int l,int r,int ql,int qr,int v)
{
    if(ql>r||qr<l)return ;
    if(ql<=l&&r<=qr) {
        lz[rt]+=v;
        push_up2(rt,l,r);
        return ;
    }
    int mid=(l+r)>>1;
    update2(rt<<1,l,mid,ql,qr,v);
    update2(rt<<1|1,mid+1,r,ql,qr,v);
    push_up2(rt,l,r);
}
int main()
{
    int lx,rx,ly,ry,n,i,j;
    scanf("%d",&n);
    for(i=0; i<n; i++) {
        scanf("%d%d%d%d",&lx,&ly,&rx,&ry);
        _hx[i]=lx,_hx[i+n]=rx,_hy[i]=ly,_hy[i+n]=ry;

        seg1[i].lx=lx,seg1[i].rx=rx,seg1[i].v=1,seg1[i].h=ly;
        seg1[i+n].lx=lx,seg1[i+n].rx=rx,seg1[i+n].v=-1,seg1[i+n].h=ry;

        seg2[i].ly=ly,seg2[i].ry=ry,seg2[i].v=1,seg2[i].h=lx;
        seg2[i+n].ly=ly,seg2[i+n].ry=ry,seg2[i+n].v=-1,seg2[i+n].h=rx;
    }
    sort(_hx,_hx+n*2);
    sort(_hy,_hy+n*2);
    sort(seg1,seg1+n*2);
    sort(seg2,seg2+n*2);
    int m1=unique(_hx,_hx+n*2)-_hx;
    int m2=unique(_hy,_hy+n*2)-_hy;
    int ans=0,ql,qr;
    memset(sum,0,sizeof(sum));
    memset(lz,0,sizeof(lz));
    for(i=0; i<n*2; i++) {
        ql=lower_bound(_hx,_hx+m1,seg1[i].lx)-_hx;
        qr=lower_bound(_hx,_hx+m1,seg1[i].rx)-_hx-1;
        int t=sum[1];
        update1(1,0,m1-1,ql,qr,seg1[i].v);
        ans+=abs(sum[1]-t);
    }
    memset(sum,0,sizeof(sum));
    memset(lz,0,sizeof(lz));
    for(i=0; i<n*2; i++) {
        ql=lower_bound(_hy,_hy+m2,seg2[i].ly)-_hy;
        qr=lower_bound(_hy,_hy+m2,seg2[i].ry)-_hy-1;
        int t=sum[1];
        update2(1,0,m2-1,ql,qr,seg2[i].v);
        ans+=abs(sum[1]-t);
    }
    printf("%d\n",ans);
    return 0;
}

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

Output

Sample Input

7
-15 0 5 10
-5 8 20 25
15 -4 24 14
0 -6 16 4
2 15 10 22
30 10 36 20
34 0 40 16

Sample Output

228

Source

POJ训练计划1177_Picture(扫描线/线段树+离散),布布扣,bubuko.com

POJ训练计划1177_Picture(扫描线/线段树+离散)

标签：des   style   http   color   os   io   strong   for   

原文地址：http://blog.csdn.net/juncoder/article/details/38521481