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POJ训练计划1177_Picture(扫描线/线段树+离散)

2024-07-17 05:57:24   220人阅读

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;
}

Picture
Time Limit: 2000MS Memory Limit: 10000K
Total Submissions: 10332 Accepted: 5485

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.

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

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

IOI 1998


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