NC51111 Atlantis

Posted 2023-05-07 空白菌

题目链接

题目

题目描述

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.

输入描述

The input consists of several test cases. Each test case starts with a line containing a single integer n \\((1 \\leq n \\leq 100)\\) of available maps. The n following lines describe one map each. Each of these lines contains four numbers \\(x_1;y_1;x_2;y_2\\) \\((0 \\leq x_1 \\lt x_2 \\leq 100000;0 \\leq y_1 \\lt y_2 \\leq 100000)\\) , not necessarily integers. The values \\((x_1; y_1)\\) and \\((x_2;y_2)\\) 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.

输出描述

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.

示例1

输入

2
10 10 20 20
15 15 25 25.5
0

输出

Test case #1
Total explored area: 180.00

题解

知识点：扫描线，线段树，离散化。

线段树+扫面线处理面积并问题，是板子题。

更新时，通过到上次更新的距离与线段覆盖长度，来计算面积。

时间复杂度 \\(O(n \\log n)\\)

空间复杂度 \\(O(n)\\)

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

template<class T>
struct Discretization 
    vector<T> uniq;
    Discretization() 
    Discretization(const vector<T> &src)  init(src); 
    void init(const vector<T> &src) 
        uniq = src;
        sort(uniq.begin() + 1, uniq.end());
        uniq.erase(unique(uniq.begin() + 1, uniq.end()), uniq.end());
    
    int get(T x)  return lower_bound(uniq.begin() + 1, uniq.end(), x) - uniq.begin(); 
;

template<class T>
class ScanlineA 
    struct Segment 
        int l, r;
        int cover;
        T len;
    ;

    int n;
    vector<T> dot;
    vector<Segment> node;

    void push_up(int rt) 
        if (node[rt].cover) node[rt].len = dot[node[rt].r + 1] - dot[node[rt].l];
        else if (node[rt].l == node[rt].r) node[rt].len = 0;
        else node[rt].len = node[rt << 1].len + node[rt << 1 | 1].len;
    

    void update(int rt, int l, int r, int x, int y, int cover) 
        if (r < x || y < l) return;
        if (x <= l && r <= y) return node[rt].cover += cover, push_up(rt);
        int mid = l + r >> 1;
        update(rt << 1, l, mid, x, y, cover);
        update(rt << 1 | 1, mid + 1, r, x, y, cover);
        push_up(rt);
    

public:
    ScanlineA() 
    ScanlineA(const vector<T> &_dot)  init(_dot); 
    void init(const vector<T> &_dot) 
        assert(_dot.size() >= 2);
        n = _dot.size() - 2;
        dot = _dot;
        node.assign(n << 2,  0,0,0,0 );
        function<void(int, int, int)> build = [&](int rt, int l, int r) 
            node[rt] =  l,r,0,0 ;
            if (l == r) return;
            int mid = l + r >> 1;
            build(rt << 1, l, mid);
            build(rt << 1 | 1, mid + 1, r);
        ;
        build(1, 1, n);
    

    void update(int x, int y, int cover)  update(1, 1, n, x, y, cover); 

    Segment query()  return node[1]; 
;
/// 面积并扫描线特化线段树，O(logn)，配合离散化可以处理任意精度覆盖长度并问题
/// 求面积并，O(nlogn)，面积并 = sum(两次扫描的距离*覆盖长度并)
//* 其中n代表线段数，并非端点数，端点数应为n+1
//* 端点编号从1开始，线段编号也从1开始
//* 任何区间（如l,r或x,y）都代表线段编号而非端点编号，即表示dot[l]到dot[r + 1]，使用时注意

template<class T>
struct pk 
    T val;
    friend bool operator<(const pk &a, const pk &b) 
        if (abs(a.val - b.val) < 1e-6) return false;//! 浮点型注意相等条件
        return a.val < b.val;
    
    friend bool operator==(const pk &a, const pk &b)  return !(a < b) && !(b < a); 
;
//* 专门处理浮点型比较判断的封装类

template<class T>
struct edge 
    T x;
    pk<T> y1, y2;
    int rky1, rky2;
    int flag;
;

int main() 
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    int cnt = 0;
    cout << fixed << setprecision(2);
    while (cnt++, cin >> n, n) 
        if (cnt > 1) cout << \'\\n\';
        vector<edge<double>> e(2 * n + 1);
        vector<pk<double>> y_src(2 * n + 1);
        for (int i = 1;i <= n;i++) 
            double x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;
            e[2 * i - 1] =  x1,y1,y2,0,0,1 ;
            e[2 * i] =  x2,y1,y2,0,0,-1 ;
            y_src[2 * i - 1] =  y1 ;
            y_src[2 * i] =  y2 ;
        

        Discretization<pk<double>> dc(y_src);
        for (int i = 1;i <= n;i++) 
            e[2 * i - 1].rky1 = dc.get( e[2 * i - 1].y1 );
            e[2 * i - 1].rky2 = dc.get( e[2 * i - 1].y2 );
            e[2 * i].rky1 = dc.get( e[2 * i].y1 );
            e[2 * i].rky2 = dc.get( e[2 * i].y2 );
        
        sort(e.begin() + 1, e.end(), [&](const auto &a, const auto &b) return a.x < b.x;);

        vector<double> dot(dc.uniq.size());
        for (int i = 1;i < dot.size();i++) dot[i] = dc.uniq[i].val;
        ScanlineA<double> sla(dot);
        double ans = 0;
        sla.update(e[1].rky1, e[1].rky2 - 1, e[1].flag);
        for (int i = 2;i <= 2 * n;i++) 
            ans += (e[i].x - e[i - 1].x) * sla.query().len;
            sla.update(e[i].rky1, e[i].rky2 - 1, e[i].flag);
        
        cout << "Test case #" << cnt << \'\\n\';
        cout << "Total explored area: " << ans << \'\\n\';
    
    return 0;

/*
2
10 10 20 20
15 15 25 25.5
2
10 10 20 20
15 15 25 25.5
0

Test case #1
Total explored area: 180.00

Test case #2
Total explored area: 180.00
 */

本文来自博客园，作者：空白菌，转载请注明原文链接：https://www.cnblogs.com/BlankYang/p/17378729.html

线段树Atlantis

Atlantis

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23181		Accepted: 8644

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 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

 

题目大意：给出一些矩形，算他们的面积并

试题分析：乍一看并不是用线段树，知道扫描线后就很好做了。

　　　　　扫描线是什么呢？

　　　　扫描线可以从左到右或者从上到下（方向无所谓），就是从上开始扫，扫到一个矩形的开头那么这个矩形就可以进入我们的面积计算，扫到末尾就退出。

　　　　这一过程我们就可以用线段树来维护，由于坐标可能有浮点数，所以就需要操作时离散化一下。

　　　　这次没有写差分+离散化+线段树，差分要标记下传，直接储存也挺好做的

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c==\'-\') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-\'0\';
	return x*f;
}
const int MAXN=100001;
const int INF=999999;
int N,M;
double a[100001],b[100001];
int A[100001];
struct data2{
	double sum,len;
	int ct;
}tr[1000001];
int tmp,tmp2;
struct data{
	double x,y1,y2;
	bool end;
}poi[1000001];
int T;

bool cmp(data a,data b){
	return a.x<b.x;
}
void build(int l,int r,int rt){
	tr[rt].ct=0;tr[rt].sum=0;
	tr[rt].len=a[r]-a[l];
	if(l+1>=r) return ;
	int mid=(l+r)>>1; 
	build(l,mid,rt*2);
	build(mid,r,rt*2+1);
	return ;
}
void Update(int rt,int l,int r){
	if(tr[rt].ct) tr[rt].sum=tr[rt].len;
	else if(r-l>1) tr[rt].sum=tr[rt*2].sum+tr[rt*2+1].sum;
	else tr[rt].sum=0;
	return ;
}
void add(int l,int r,int rt,int L,int R){
	if(L==l&&R==r){
		tr[rt].ct++;
		Update(rt,l,r);
		return ;
	}
	int mid=(l+r)>>1;
	if(L>=mid) add(mid,r,rt*2+1,L,R);
	else if(R<=mid) add(l,mid,rt*2,L,R);
	else{
		add(mid,r,rt*2+1,mid,R);
		add(l,mid,rt*2,L,mid);
	}
	Update(rt,l,r);
	return;
}
void del(int l,int r,int rt,int L,int R){
	if(L==l&&R==r){
		tr[rt].ct--;
		Update(rt,l,r);
		return ;
	}
	int mid=(l+r)>>1;
	if(L>=mid) del(mid,r,rt*2+1,L,R);
	else if(R<=mid) del(l,mid,rt*2,L,R);
	else{
		del(mid,r,rt*2+1,mid,R);
		del(l,mid,rt*2,L,mid);
	}
	Update(rt,l,r);
	return;
}

double ans;
int main(){
    while(1){
    	N=read();
    	if(!N) break;
    	tmp2=tmp=0;double x1,x2,y1,y2;
		for(int i=1;i<=N;i++){
			cin>>x1>>y1>>x2>>y2;
			poi[++tmp].x=x1;
			poi[tmp].y1=y1;
			poi[tmp].y2=y2;
			poi[tmp].end=false;
			poi[++tmp].x=x2;
			poi[tmp].y1=y1;
			poi[tmp].y2=y2;
			poi[tmp].end=true;
		}
		for(int i=1;i<=tmp;i+=2){
			a[tmp2]=b[tmp2]=poi[i].y1;tmp2++;
		    a[tmp2]=b[tmp2]=poi[i].y2;tmp2++;
		} 
		sort(poi+1,poi+tmp+1,cmp);
		sort(a,a+tmp2);
		int tmp3=0,tmp4=0;
		tmp3=unique(a,a+tmp2)-a;
		ans=0;
		build(0,tmp3-1,1);
		double lasth;
		for(int i=1;i<tmp;i++){
			if(!poi[i].end) add(0,tmp3-1,1,lower_bound(a,a+tmp3,poi[i].y1)-a,lower_bound(a,a+tmp3,poi[i].y2)-a);
			else del(0,tmp3-1,1,lower_bound(a,a+tmp3,poi[i].y1)-a,lower_bound(a,a+tmp3,poi[i].y2)-a);
			ans+=(double)((poi[i+1].x-poi[i].x)*tr[1].sum);
		}
		printf("Test case #%d\\nTotal explored area: %.2f\\n\\n",++T,ans);
	} 
}