线段树扫描线——HDU 1542

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Atlantis

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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

题意：矩阵面积并

思路：线段树扫描线，从左往右扫或从下往上扫，含浮点数，故要离散化

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=2000+10;
#define ms(x,y) memset(x,y,sizeof(x))
using namespace std;
double x[MAXN];
double sum[MAXN<<2];
int cnt[MAXN<<2];

struct Line 
{
	double y,x1,x2;
	int flag;
}line_y[MAXN];

bool cmp(Line l1, Line l2)
{
	return l1.y < l2.y;
}

void up(int rt, int left, int right)
{
	if(cnt[rt]) sum[rt] = x[right+1] - x[left];
	else if(left == right) sum[rt] = 0;
	else sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void update(int rt, int left, int right, int l, int r, int flag)
{
	if(l == left && right == r){
		cnt[rt]+=flag;
		up(rt, left, right);
		return;
	}
	int mid=(left + right)>>1;
	if(mid >= r) update(rt<<1, left, mid, l, r, flag);
	else if(mid < l) update(rt<<1|1, mid+1, right, l, r, flag);
	else{
		 update(rt<<1, left, mid, l, mid, flag);
		 update(rt<<1|1, mid+1, right, mid+1, r, flag);
	}
	up(rt, left, right);
}

int binary(double key, int n, double  *a)
{
	int l=0, r=n-1;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(a[mid] == key) return mid;
		if(a[mid] > key) r = mid - 1;
		else l = mid + 1;
	}
	return -1;
}

int main()
{
	//freopen("in.txt","r",stdin);
	int n,w=0;
	while(scanf("%d", &n), n)
	{
		ms(x,0);
		ms(cnt,0);
		ms(sum,0);
		printf("Test case #%d\n", ++w);
		int u=0;
		int n_x=0;
		for(int i=0; i<n; i++){
			double x1,y1,x2,y2;
			scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
			x[n_x++]=x1;
			x[n_x++]=x2;
			line_y[u].y=y1;
			line_y[u].flag=1;
			line_y[u].x1=x1;
			line_y[u++].x2=x2;
			line_y[u].y=y2;
			line_y[u].flag=-1;
			line_y[u].x1=x1;
			line_y[u++].x2=x2;
		}
		sort(x, x+n_x);//横坐标离散化
		sort(line_y, line_y+u, cmp);//扫描线排序
		int k=1;
		for(int i=1; i<n_x; i++) if(x[i] != x[i-1]) x[k++] = x[i];
		double ans=0.0;
		for(int i=0; i<u-1; i++){
			int l=binary(line_y[i].x1, k, x);
			int r=binary(line_y[i].x2, k, x);
			update(1, 0, k-2, l, r-1, line_y[i].flag);
			ans+= sum[1]*(line_y[i+1].y - line_y[i].y);
		}
		printf("Total explored area: %.2lf\n\n", ans);
	}
	return 0;
}