POJ 3277 City Horizon

最新推荐文章于 2019-06-07 21:19:27 发布

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本文介绍了一种使用线段树数据结构解决计算多个矩形建筑组成的天际线总面积问题的方法。通过输入建筑物的位置和高度信息，算法能够高效地计算出不重叠区域的总面积。

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City Horizon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15769		Accepted: 4276

Description

Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings.

The entire horizon is represented by a number line with N (1 ≤ N ≤ 40,000) buildings. Building i's silhouette has a base that spans locations A_i through B_i along the horizon (1 ≤ A_i < B_i ≤ 1,000,000,000) and has height H_i (1 ≤ H_i ≤ 1,000,000,000). Determine the area, in square units, of the aggregate silhouette formed by all N buildings.

Input

Line 1: A single integer: N
Lines 2.. N+1: Input line i+1 describes building i with three space-separated integers: A_i, B_i, and H_i

Output

Line 1: The total area, in square units, of the silhouettes formed by all N buildings

Sample Input

Sample Output

Hint

The first building overlaps with the fourth building for an area of 1 square unit, so the total area is just 3*1 + 1*4 + 2*2 + 2*3 - 1 = 16.

Source

USACO 2007 Open Silver

面积并线段树

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#define N 41000
using namespace std;
int y[N*2];
struct num
{
    int x;
    int y1,y2,flag;
}a[N*2];
struct Num
{
    int l,r,tl,tr,sum,tlen;
}b[N*8];
bool cmp(num p1,num p2)
{
    return p1.x<p2.x;
}
int main()
{
    //freopen("data.txt","r",stdin);
    void build(int k,int l,int r);
    void update(int k,num p);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            int x1,x2;
            int y2;
            scanf("%d %d %d",&x1,&x2,&y2);
            y[i*2-1] = 0;
            y[i*2] = y2;
            a[i*2-1].x = x1;
            a[i*2-1].y1 = 0;
            a[i*2-1].y2 = y2;
            a[i*2-1].flag = 1;
            a[i*2].x = x2;
            a[i*2].y1 = 0;
            a[i*2].y2 = y2;
            a[i*2].flag = -1;
        }
        n  = n * 2;
        sort(y+1,y+n+1);
        sort(a+1,a+n+1,cmp);
        build(1,1,n);
        update(1,a[1]);
        __int64 s=0;
        for(int i=2;i<=n;i++)
        {
            s+=(__int64)(a[i].x-a[i-1].x)*(__int64)(b[1].tlen);
            update(1,a[i]);
        }
        printf("%I64d\n",s);
    }
    return 0;
}
void build(int k,int l,int r)
{
    b[k].l =l; b[k].r = r;
    b[k].tl = y[l];
    b[k].tr = y[r];
    b[k].sum = b[k].tlen=0;
    if(l+1==r)
    {
        return ;
    }
    int mid = (l+r)>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid,r);
}
void cal(int k)
{
    if(b[k].sum>0)
    {
        b[k].tlen = b[k].tr - b[k].tl;
        return ;
    }
    if(b[k].l+1==b[k].r)
    {
        b[k].tlen = 0;
    }else
    {
        b[k].tlen = b[k<<1].tlen+b[k<<1|1].tlen;
    }
}
void update(int k,num p)
{
    if(b[k].tl==p.y1&&b[k].tr==p.y2)
    {
        b[k].sum+=p.flag;
        cal(k);
        return ;
    }
    if(b[k<<1].tr>=p.y2)
    {
        update(k<<1,p);
    }else if(b[k<<1|1].tl<=p.y1)
    {
        update(k<<1|1,p);
    }else
    {
        num p1 = p;
        p1.y2 = b[k<<1].tr;
        update(k<<1,p1);
        p1 = p;
        p1.y1 = b[k<<1|1].tl;
        update(k<<1|1,p1);
    }
    cal(k);
}