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(中等) HDU 3265 Posters , 扫描线。

时间:2015-01-11 22:46:50      阅读:431      评论:0      收藏:0      [点我收藏+]

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  Problem Description
  Ted has a new house with a huge window. In this big summer, Ted decides to decorate the window with some posters to prevent the glare outside. All things that Ted can find are rectangle posters. 

  However, Ted is such a picky guy that in every poster he finds something ugly. So before he pastes a poster on the window, he cuts a rectangular hole on that poster to remove the ugly part. Ted is also a careless guy so that some of the pasted posters may overlap when he pastes them on the window. 

  Ted wants to know the total area of the window covered by posters. Now it is your job to figure it out.

  To make your job easier, we assume that the window is a rectangle located in a rectangular coordinate system. The window’s bottom-left corner is at position (0, 0) and top-right corner is at position (50000, 50000). The edges of the window, the edges of the posters and the edges of the holes on the posters are all parallel with the coordinate axes. 
 
  题目就是求矩形的并的面积,典型的扫描线的题目,对于中间有个洞的矩形,可以拆分成四个来做。
  技术分享
  网上还有一种办法,就是对扫描线的线段树进行改进,在HDU这个题的Discuss里面有。
  
代码如下:
技术分享
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define lc po*2
#define rc po*2+1
#define lson L,M,lc
#define rson M+1,R,rc

using namespace std;

struct BIAN
{
    int x,y1,y2;
    short state;
};

const int maxn=50005;

BIAN bian[50005*8];
int BIT[maxn*4];
int COL[maxn*4];

void pushUP(int L,int R,int po)
{
    if(COL[po])
        BIT[po]=R+1-L;
    else if(L==R)
        BIT[po]=0;
    else
        BIT[po]=BIT[lc]+BIT[rc];
}

void update(int ul,int ur,int ut,int L,int R,int po)
{
    if(ul<=L&&ur>=R)
    {
        COL[po]+=ut;
        pushUP(L,R,po);

        return;
    }

    int M=(L+R)/2;

    if(ul<=M)
        update(ul,ur,ut,lson);
    if(ur>M)
        update(ul,ur,ut,rson);

    pushUP(L,R,po);
}

bool cmp(BIAN a,BIAN b)
{
    return a.x<b.x;
}

int main()
{
    int N;
    long long ans;
    int x1,x2,x3,x4,y1,y2,y3,y4;

    while(~scanf("%d",&N))
    {
        if(!N)
            break;

        memset(COL,0,sizeof(COL));
        memset(BIT,0,sizeof(BIT));
        ans=0;

        for(int i=1;i<=N;++i)
        {
            scanf("%d %d %d %d %d %d %d %d",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);

            bian[i*8-5].x=x1;
            bian[i*8-5].y1=y1;
            bian[i*8-5].y2=y2;
            bian[i*8-5].state=1;
            
            bian[i*8-4].x=x3;
            bian[i*8-4].y1=y1;
            bian[i*8-4].y2=y2;
            bian[i*8-4].state=-1;

            bian[i*8-3].x=x4;
            bian[i*8-3].y1=y1;
            bian[i*8-3].y2=y2;
            bian[i*8-3].state=1;

            bian[i*8-2].x=x2;
            bian[i*8-2].y1=y1;
            bian[i*8-2].y2=y2;
            bian[i*8-2].state=-1;

            bian[i*8-1].x=x3;
            bian[i*8-1].y1=y4;
            bian[i*8-1].y2=y2;
            bian[i*8-1].state=1;

            bian[i*8].x=x4;
            bian[i*8].y1=y4;
            bian[i*8].y2=y2;
            bian[i*8].state=-1;

            bian[i*8-7].x=x3;
            bian[i*8-7].y1=y1;
            bian[i*8-7].y2=y3;
            bian[i*8-7].state=1;

            bian[i*8-6].x=x4;
            bian[i*8-6].y1=y1;
            bian[i*8-6].y2=y3;
            bian[i*8-6].state=-1;
        }

        sort(bian+1,bian+8*N+1,cmp);

        for(int i=1;i<=8*N;++i)
        {
            ans+=(long long)BIT[1]*(bian[i].x-bian[i-1].x);
            
            if(bian[i].y2>bian[i].y1)
                update(bian[i].y1,bian[i].y2-1,bian[i].state,0,50000,1);
        }

        cout<<ans<<endl;
    }

    return 0;
}
View Code

 

(中等) HDU 3265 Posters , 扫描线。

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原文地址:http://www.cnblogs.com/whywhy/p/4217137.html

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