HDU 1542 扫描线

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16397    Accepted Submission(s): 6691

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

 

题意：在一个二位平面上放置矩形，给出矩形的左下角坐标和右上角坐标，求整个二位平面的覆盖面积

思路：扫描线

先看一张我手画的样例草图

看上面的图，我做的扫面线是横线（与X轴平行），从下往上扫描

存进一个数组里从小到大排序，用下标做离散化

所以依次扫描的线是 1 线 -> 2线 -> 3线 -> 4线

每条线存了 两个x值（线段的左右端点），一个y值（线段高度），还有一个flag标记（1为矩形的下线，-1为上线）

扫描第一条线的时候，在下标数组中找到这条线的左右端点值所在的下标，然后update进线段树中

第一条线插入完毕后 tree[1].sum = 10  (sum表示整个区间的有效线段长度）

所以面积 +=  2线与1线的高度差 * tree[1].sum = 5 * 10 = 50

此时算出了 1 面 的面积

扫描第二条线的时候，它是一条下线，会把区间长度更新为 15，注意，获取长度的操作是利用线段树中线段端点差值来获取的

而不是通过插入的线段的值，所以更新之后获取的长度是 15，而不是叠加的20。

面积 += 2线与3线的高度差 * tree[1].sum = 75

扫描第三条线的时候发现是 上线，更新完区间之后，有效长度为 10

计算出最后一块面积55

所以总面积是 180

最需要注意的就是区间问题，一般的线段树区间是分成 [1,2][3,4][4,5]这样子的，但是这样会丢失区间

比如我们需要 [2,3]的时候就找不到这个区间了，所以扫面线的线段树需要改造

我们把 [0,0] 表示为线段 [0,1]   [1,1]表示线段 [1,2]    [0,3]表示线段 [0,4]

所以在插入线段的时候 要插入  ( l , r -1 ) , 计算区间的时候 用 pos[r + 1] - pos[l] 就可以了

贴上代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

#define maxn 100005
#define mem(a,x) memset(a,x,sizeof(a))
#define debug printf("********");
struct line{
	double l,r,h;
	int flag;
	line(){}
	line(double ll,double rr,double hh,int fflag)
	 : l(ll),r(rr),h(hh),flag(fflag){}
}lines[maxn];

struct node{
	int l,r,date;
	double sum;
}tree[5000];

int lazy[maxn * 4];
double pos[maxn];

bool cmp(line a,line b){
	return a.h < b.h;
}

void push_up(int root){
	if(lazy[root] > 0){
		tree[root].sum = pos[tree[root].r + 1] - pos[tree[root].l];
	}	
	else if(tree[root].l == tree[root].r){
		tree[root].sum = 0;
	}else{
		tree[root].sum = tree[root << 1].sum + tree[root << 1 | 1].sum;
	}
}

void build_tree(int root,int l,int r){
	tree[root].l = l;
	tree[root].r = r;
	tree[root].sum = 0;
	if(l == r)
		return ;
	int mid = (l + r) >> 1;
	build_tree(root << 1,l,mid);
	build_tree(root << 1 | 1,mid + 1,r);
}

void update(int root,int l,int r,int val){
	if(tree[root].l >= l && tree[root].r <= r){
		lazy[root] += val;
		push_up(root);
		return ;
	}
	int mid = (tree[root].l + tree[root].r) >> 1;
	if(l <= mid)
		update(root << 1,l,r,val);
	if(r > mid)
		update(root << 1 | 1,l,r,val);
	push_up(root);
}

int discre(int n){
	int tmp = 1;
	for(int i = 1;i < n;i++){
		if(pos[i] != pos[i - 1]){
			pos[tmp++] = pos[i];
		}
	}
	return tmp;
}
int main(){
	int n,cnt,num,cas = 1;
	double x1,y1,x2,y2;
	while(scanf("%d",&n) != EOF){
		if(n == 0)
			break;
		cnt = num = 0;
		mem(lazy,0);
		for(int i = 1;i <= n;i++){
			scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
			lines[cnt++] = line(x1,x2,y1,1);
			lines[cnt++] = line(x1,x2,y2,-1); 
			pos[num++] = x1;
			pos[num++] = x2; 
		}
		sort(lines,lines + cnt,cmp);
		sort(pos,pos + num);
		num = discre(num);
		double area = 0.0;
		build_tree(1,0,num - 1);
		for(int i = 0;i < cnt - 1;i++){
			int l = lower_bound(pos,pos + num,lines[i].l) - pos;
			int r = lower_bound(pos,pos + num,lines[i].r) - pos - 1;
			update(1,l,r,lines[i].flag);
			area += (lines[i + 1].h - lines[i].h) * tree[1].sum;
		}
		printf("Test case #%d\n",cas++);
		printf("Total explored area: %.2f\n\n",area);
	}
}