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 Ai through Bi along the horizon (1 ≤ Ai < Bi ≤ 1,000,000,000) and has height Hi (1 ≤ Hi ≤ 1,000,000,000). Determine the area, in square units, of the aggregate silhouette formed by all N buildings.
Input
Lines 2.. N+1: Input line i+1 describes building i with three space-separated integers: Ai, Bi, and Hi
Output
Sample Input
4 2 5 1 9 10 4 6 8 2 4 6 3
Sample Output
16
Hint
Source
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;
}