poj3277--City Horizon(线段树+离散化+扫描线)

最新推荐文章于 2020-02-12 13:06:55 发布

刀刀狗0102

最新推荐文章于 2020-02-12 13:06:55 发布

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分类专栏：数据结构计算几何

本文链接：https://blog.youkuaiyun.com/winddreams/article/details/38498597

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本文介绍了一种使用离散化和区间更新的方法来计算多个矩形建筑轮廓所形成的总阴影面积的技术。通过按x轴从左到右扫描并更新树状结构，避免了计算重叠部分的复杂性。

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City Horizon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16206		Accepted: 4414

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.

和poj1151相同，求解矩形的并的面积 http://blog.youkuaiyun.com/winddreams/article/details/38495093

按x轴有左向右扫描，对y值进行离散化，为了将每个段连接起来，所以左右子树应该是[l,mid],[mid,r]，每段的差是1，不是0，这样可以防止，忽略掉mid+1到mid这一段，然后判断，只有lazy大于0时，这一个节点的所有长度都加进去，不然节点大小 = 左右子树的和

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxn 81000
#define LL __int64
struct node1
{
    LL x, y1 , y2 ;
    LL flag ;
} p[maxn<<2];
struct node2
{
    LL l , r ;
    LL sum ;
} cl[maxn<<2];
LL lazy[maxn<<2];
LL s[maxn<<2] ;
bool cmp(node1 a,node1 b)
{
    return a.x < b.x ;
}
void push_up(int rt)
{
    if( lazy[rt] )
        cl[rt].sum = cl[rt].r - cl[rt].l ;
    else
        cl[rt].sum = cl[rt<<1].sum + cl[rt<<1|1].sum ;
}
void creat(int rt,int l,int r)
{
    cl[rt].l = s[l] ;
    cl[rt].r = s[r] ;
    if( r - l > 1 )
    {
        creat(rt<<1,l,(l+r)/2);
        creat(rt<<1|1,(l+r)/2,r);
    }
    else
        cl[rt].sum = 0 ;
    return ;
}
void update(int rt,LL l,LL r,int flag)
{
    if( l == cl[rt].l && r == cl[rt].r )
    {
        lazy[rt] += flag ;
        push_up(rt);
        return ;
    }
    else
    {
        if( cl[rt<<1].r > l )
        {
            LL k = cl[rt<<1].r < r ? cl[rt<<1].r : r ;
            update(rt<<1,l,k,flag);
        }
        if( cl[rt<<1|1].l < r )
        {
            LL k = cl[rt<<1|1].l > l ? cl[rt<<1|1].l : l ;
            update(rt<<1|1,k,r,flag);
        }
        push_up(rt);
    }
}
int main()
{
    LL i , n , x1 , x2 , h , low , ans ;
    while(scanf("%I64d", &n)!=EOF)
    {
        for(i = 0 ; i < n ; i++)
        {
            scanf("%I64d %I64d %I64d", &x1, &x2, &h);
            p[i].x = x1 ;
            p[i].y1 = 0 ;
            p[i].y2 = h ;
            p[i].flag = 1 ;
            p[i+n].x = x2 ;
            p[i+n].y1 = 0 ;
            p[i+n].y2 = h ;
            p[i+n].flag = -1 ;
            s[i+1] = h ;
        }
        s[n+1] = 0 ;
        sort(s+1,s+(n+2));
        sort(p,p+2*n,cmp);
        creat(1,1,n+1);
        update(1,p[0].y1,p[0].y2,p[0].flag);
        ans = 0 ;
        for(i = 1 ; i < 2*n ; i++)
        {
            ans += ( p[i].x - p[i-1].x )*cl[1].sum;
            update(1,p[i].y1,p[i].y2,p[i].flag);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}