hdu 1542 Atlantis(段树&扫描线&面积和)

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6386    Accepted Submission(s): 2814

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

 

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 题意：

给你一些矩形和坐标轴平行的矩形。告诉你们他们左下角和右上角的坐标。如今问你。整个坐标面被他们覆盖的面积。

思路：

因为直接找的线段树扫描线的题目。所以也没想其他思路。讲一下为什么能够用扫描线来做吧。首先是面积的求解。

面积=底边长*高。这底边长不一定要求连续。所以假设我们知道某一时刻底边长和高就能够算出面积了。而线段树就能够出色的完毕求出某个时间底边长(x轴被覆盖的长度)。

所以我们仅仅须要把全部平行与x轴的边按高度排序。然后依次插入到线段树中。遇到矩形的下边就增加到线段树中。遇到上边就把相应的下边从线段树中删除。文字可能不是非常好理解。绘图看看就知道了。

具体见代码：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=250;
#define lson L,mid,ls
#define rson mid+1,R,rs
int n,m;
int cov[maxn<<2];
double len[maxn<<2],H[maxn];
struct node
{
    double x1,x2,h;
    int v;
    node(double a=0,double b=0,double c=0,int d=0):x1(a),x2(b),h(c),v(d){}
} seg[maxn];
bool cmp(node a,node b)
{
    return a.h<b.h;
}
void init()
{
    sort(H,H+m);
    m=unique(H,H+m)-H;
}
int Hash(double x)
{
    return lower_bound(H,H+m,x)-H;
}
void PushUp(int L,int R,int rt)
{
    if(cov[rt])//有标记肯定整块覆盖了
        len[rt]=H[R+1]-H[L];
    else if(L==R)//没有左右儿子了。

        len[rt]=0;
    else//没有整块覆盖可是被部分覆盖了
        len[rt]=len[rt<<1]+len[rt<<1|1];
}
void update(int L,int R,int rt,int l,int r,int d)
{
    if(l<=L&&R<=r)
    {
        cov[rt]+=d;
        PushUp(L,R,rt);
        return;
    }//这样的标记是不用下传的。由于没删除一个上边。一定有一个下边与之相应。

    int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
    if(l<=mid)
        update(lson,l,r,d);
    if(r>mid)
        update(rson,l,r,d);
    PushUp(L,R,rt);
}
int main()
{
    int cas=1,i,ptr;
    double x1,x2,y1,y2,ans;
    while(scanf("%d",&n),n)
    {
        printf("Test case #%d
",cas++);
        for(i=ptr=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            H[ptr]=x1;
            seg[ptr++]=node(x1,x2,y1,1);
            H[ptr]=x2;
            seg[ptr++]=node(x1,x2,y2,-1);
        }
        m=ptr,ans=0;
        init();
        sort(seg,seg+ptr,cmp);
        memset(len,0,sizeof len);
        memset(cov,0,sizeof cov);
        for(i=0,ptr--;i<ptr;i++)
        {
            update(0,m-1,1,Hash(seg[i].x1),Hash(seg[i].x2)-1,seg[i].v);//m个结点m-1条线段
            ans+=(seg[i+1].h-seg[i].h)*len[1];
        }
        printf("Total explored area: %.2lf

",ans);
    }
    return 0;
}

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