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HDU 1542 Atlantis(线段树扫描线)

2024-07-15 13:14:26   222人阅读

http://acm.hdu.edu.cn/showproblem.php?pid=1542

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6788    Accepted Submission(s): 2970


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
 


题意：

给出N个矩形，边与坐标轴平行，问最后组成图形的面积

分析：

线段树扫描线。

因为坐标是浮点数，而且有点大，所以先要离散化，我是用平行于X轴的线从下往上扫的，所以只需要离散化X坐标;前几天多校某一题的标程给了一个离散化的好方法：使用sort() & unique()，我之前都是用map(红黑树);

扫描线要怎么扫呢？首先要把矩形平行于X轴的边分出来，分为下边(1)和上边(-1)，并且记录高度(y)，然后以高度升序排列;然后就开始扫，遇到下边对应线段+1,上边就-1,这样大于0的区间就对应有线段，每次扫描就计算出对应的面积，求和即可。

维护左闭右开区间的优势显现出来了，哈哈哈哈哈哈哈。。。。



如上图加入第一根线



如上图计算出蓝色面积



如上图加入第二根线



如上图计算绿色部分面积



如上图加入第三根线



如上图计算紫色部分面积



如上图加入最后一根线(that‘s all)


#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm
#define maxn 202

using namespace std;

struct EDGE
{
    double xl,xr,y;
    int v;
    void _set(double a,double b,double h,double val)
    {
        xl=a;xr=b;y=h;v=val;
    }
    bool operator < (const EDGE &e)const
    {
        return y<e.y;
    }
}edge[202];

double X[maxn];
double sum[maxn<<2];
int val[maxn<<2];

inline void pushup(int k,int l,int r)
{
    if (val[k])
        sum[k]=X[r]-X[l];
    else    if (r-l==1)
                sum[k]=0;
            else
                sum[k]=sum[k*2+1]+sum[k*2+2];
}

void update(int a,int b,int v,int k,int l,int r)
{
    if (b<=l || r<=a)   return ;
    if (a<=l && r<=b)
    {
        val[k]+=v;
        pushup(k,l,r);
    }
    else
    {
        update(a,b,v,k*2+1,l,l+r>>1);
        update(a,b,v,k*2+2,l+r>>1,r);
        pushup(k,l,r);
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("/home/fcbruce/文档/code/t","r",stdin);
    #endif // ONLINE_JUDGE

    int n;
    double x1,x2,y1,y2;
    int m=202;
    int T_T=0;

    while (scanf("%d",&n),n)
    {
        for (int i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            edge[i]._set(x1,x2,y1,1);
            edge[i+n]._set(x1,x2,y2,-1);
            X[i]=x1;X[i+n]=x2;
        }
        n<<=1;
        sort(edge,edge+n);
        sort(X,X+n);
        int xn=unique(X,X+n)-X;
        double area=0;

        for (int i=0;i<n;i++)
        {
            if (i)  area+=(edge[i].y-edge[i-1].y)*sum[0];
            int a=lower_bound(X,X+xn,edge[i].xl)-X;
            int b=lower_bound(X,X+xn,edge[i].xr)-X;
            int v=edge[i].v;
            update(a,b,v,0,0,m);
        }

        T_T++;
        printf("Test case #%d\n",T_T);
        printf("Total explored area: %.2f\n\n",area);
    }

    return 0;
}



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