HDU 1542 Atlantis (線段樹+離散化+掃描線)

Mr_Treeeee發表於2020-04-06


Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13446    Accepted Submission(s): 5588


Problem Description
There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.
 

Input
The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.

The input file is terminated by a line containing a single 0. Don’t process it.
 

Output
For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.

Output a blank line after each test case.
 

Sample Input
2
10 10 20 20
15 15 25 25.5
0





 



Sample Output




Test case #1
Total explored area: 180.00 





 



Source



Mid-Central European Regional Contest 2000 



 




題意：



給你n個矩形，讓你算出總面積，重疊的當然不能算。





POINT：


離散化不說了。


掃描線法，具體看了很多大神的部落格，可以去搜搜。


下面這兩段要好好理解一下。





if(cnt[x])sum[x]=hah[r+1]-hah[l];





R=lower_bound(hah+1,hah+1+m,len[i].r)-hah-1



可以想象成每個點都是一條線段，區間(1,1)代表端點1到2的長度，區間(1,2)代表端點1到3的長度。



為什麼這樣呢，走過彎路除錯一下就知道了呀！直接聽懂不是很無趣嗎？（你只是不會講而已別說了


#include <iostream>
#include <stdio.h>
#include <math.h>
#include <map>
#include <string.h>
#include <algorithm>
#include <vector>
#include <fstream>
using namespace std;
#define lt 2*x
#define rt 2*x+1
#define LL long long
const int N = 210*10;
struct node
{
    double l,r;
    int flag;
    double h;
}len[N/10];
int cnt[N];
double sum[N];
double hah[N/10];
bool cmd(node a,node b)
{
    return a.h<b.h;
}
void init()
{
    memset(hah,0,sizeof hah);
    memset(len,0,sizeof len);
    memset(sum,0,sizeof sum);
    memset(cnt,0,sizeof cnt);
}
void pushup(int x,int l,int r)
{
    if(cnt[x]) sum[x]=hah[r+1]-hah[l];
    else if(l==r) sum[x]=0;
    else sum[x]=sum[rt]+sum[lt];
}
void add(int x,int l,int r,int ll,int rr,int flag)
{
    if(ll<=l&&rr>=r)
    {
        cnt[x]+=flag;
    }
    else
    {
        int mid=(l+r)>>1;
        if(ll<=mid) add(lt,l,mid,ll,rr,flag);
        if(mid<rr) add(rt,mid+1,r,ll,rr,flag);
    }
    pushup(x,l,r);
}
int main()
{
    int n;
    int p=0;
    while(~scanf("%d",&n)&&n)
    {
        init();
        int num=0;
        for(int i=1;i<=n;i++)
        {
            double x1,y1,x2,y2;
            scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
            len[++num].h=y1,len[num].l=x1,len[num].r=x2;
            hah[num]=x1;
            len[num].flag=1;
            len[++num].h=y2,len[num].l=x1,len[num].r=x2;
            hah[num]=x2;
            len[num].flag=-1;
        }
        sort(len+1,len+1+num,cmd);
        sort(hah+1,hah+1+num);
        int m=1;
        for(int i=2;i<=num;i++)
            if(hah[i]!=hah[i-1]) hah[++m]=hah[i];
        double ans=0;
        for(int i=1;i<=num;i++)
        {
            int L=lower_bound(hah+1,hah+1+m,len[i].l)-hah;
            int R=lower_bound(hah+1,hah+1+m,len[i].r)-hah-1;
            add(1,1,m,L,R,len[i].flag);
            ans+=sum[1]*(len[i+1].h-len[i].h);
        }
        printf("Test case #%d\nTotal explored area: %.2lf\n\n",++p,ans);
        
    }
    return 0;
}



                                    




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