POJ3277-City Horizon_poj 3277-优快云博客

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/75196061

本文介绍了一道经典的计算几何问题——求解由多栋建筑物组成的天际线的总面积。通过坐标离散化和线段树的数据结构来解决建筑物重叠部分的计算问题。

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City Horizon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19266		Accepted: 5324

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

题意：有n栋建筑它们有不同的高度，求从一个面看进入能看到的总面积

解题思路：先将坐标离散化，然后线段树维护每一段的最高高度即可

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n;
LL l[40005], r[40005], h[40005], x[80005], a[320005];

void update(int k, int l, int r, int ll, int rr, LL val)
{
	if (ll >= r || rr <= l) return;
	if (l >= ll&&r <= rr)
	{
		a[k] = max(a[k], val);
		return;
	}
	if (a[k] >= val) return;
	int mid = (l + r) >> 1;
	update(k << 1, l, mid, ll, rr, val);
	update(k << 1 | 1, mid, r, ll, rr, val);
}

LL query(int k, int l, int r, LL h)
{
	if (h>a[k]) a[k] = h;
	if (l + 1 == r) return (x[r] - x[l])*a[k];
	int mid = (l + r) >> 1;
	return query(k << 1, l, mid, a[k]) + query(k << 1 | 1, mid, r, a[k]);
}

int main()
{
	while (~scanf("%d", &n))
	{
		int cnt = 1, ans = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%lld%lld%lld", &l[i], &r[i], &h[i]);
			x[cnt++] = l[i];
			x[cnt++] = r[i];
		}
		sort(x + 1, x + cnt);
		int m = unique(x + 1, x + cnt) - x;
		memset(a, 0, sizeof a);
		for (int i = 1; i <= n; i++)
		{
			int ll = lower_bound(x + 1, x + m, l[i]) - x;
			int rr = lower_bound(x + 1, x + m, r[i]) - x;
			update(1, 1, m - 1, ll, rr, h[i]);
		}
		printf("%lld\n", query(1, 1, m - 1, 0));
	}
	return 0;
}