POJ 题目3277 City Horizon（线段树+离散化+扫描线）_acm题目224. city horizon-优快云博客

本文介绍了一个计算城市建筑轮廓总面积的问题，通过输入每栋建筑物的位置和高度，利用离散化和线段树等数据结构和算法，高效求解重叠部分并最终得出所有建筑物轮廓所覆盖的总面积。

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City Horizon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17146		Accepted: 4668

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

数组开小wa了，几次，n=4000，离散化8000，线段树最少3倍啊

ac代码

Problem: 3277  User: kxh1995 
Memory: 7748K  Time: 735MS 
Language: C++  Result: Accepted

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct s
{
	__int64 x1,x2,y;
	int flag;
}a[380080];
__int64 hash[380050],sum[380050];
int col[380050];
int cmp(s a,s b)
{
	return a.y<b.y;
}
__int64 bseach(__int64 key,__int64 n)
{
	__int64 l=1,r=n;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(hash[mid]==key)
			return mid;
		if(hash[mid]<key)
			l=mid+1;
		else
			r=mid-1;
	}
	return l;
}
void pushup(__int64 tr,__int64 l,__int64 r)
{
	if(col[tr])
		sum[tr]=hash[r+1]-hash[l];
	else
		if(l==r)
			sum[tr]=0;
		else
			sum[tr]=sum[tr<<1]+sum[tr<<1|1];
}
void update(__int64 L,__int64 R,__int64 l,__int64 r,__int64 tr,__int64 flag)
{
	if(L<=l&&r<=R)
	{
		col[tr]+=flag;
		pushup(tr,l,r);
		return;
	}
	int mid=(l+r)>>1;
	if(L<=mid)
		update(L,R,l,mid,tr<<1,flag);
	if(R>mid)
		update(L,R,mid+1,r,tr<<1|1,flag);
	pushup(tr,l,r);
}
int main()
{
	__int64 n;
	while(scanf("%I64d",&n)!=EOF)
	{
		memset(col,0,sizeof(col));
		memset(sum,0,sizeof(sum));
		__int64 x1,x2,y;
		__int64 i,k=1;
		for(i=1;i<=n;i++)
		{
			scanf("%I64d%I64d%I64d",&x1,&x2,&y);
			a[k].x1=x1;
			a[k].x2=x2;
			a[k].y=1;
			a[k].flag=1;
			hash[k++]=x1;
			a[k].x1=x1;
			a[k].x2=x2;
			a[k].y=y+1;
			a[k].flag=-1;
			hash[k++]=x2;
		}
		sort(a+1,a+k,cmp);
		sort(hash+1,hash+k);
		__int64 ans=0;
		for(i=1;i<k-1;i++)
		{
			__int64 x,y;
			x=bseach(a[i].x1,k-1);
			y=bseach(a[i].x2,k-1)-1;
			if(x<=y)
				update(x,y,1,k-1,1,a[i].flag);
			ans+=sum[1]*(a[i+1].y-a[i].y);
		}
		printf("%I64d\n",ans);
	}
}