[POJ3277]City Horizon

试题描述

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.

输入

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

输出

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

输入示例

4
2 5 1
9 10 4
6 8 2
4 6 3

输出示例

16

数据规模及约定
见“试题描述”

题解

上一题的代码改一改就好了。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
	return x * f;
}

#define maxn 40010
#define LL long long
int n, cnt, ca, cd;
struct Rec {
	LL x1, y1, x2, y2;
	Rec() {}
	Rec(LL _1, LL _2, LL _3, LL _4): x1(_1), y1(_2), x2(_3), y2(_4) {}
} rs[maxn];
struct Rec_int { int x1, y1, x2, y2; } rsi[maxn];
LL num[maxn<<1], A[maxn<<1], B[maxn<<1];
struct Line {
	int l, r, x;
	Line() {}
	Line(int _1, int _2, int _3): l(_1), r(_2), x(_3) {}
	bool operator < (const Line& t) const { return x < t.x; }
} ad[maxn], de[maxn];

LL sumv[maxn<<3];
int addv[maxn<<3];
void build(int L, int R, int o) {
	if(L == R) {
		sumv[o] = 0;
		addv[o] = 0;
		return ;
	}
	int M = L + R >> 1, lc = o << 1, rc = lc | 1;
	build(L, M, lc); build(M+1, R, rc);
	sumv[o] = 0; addv[o] = 0;
	return ;
}
void update(int L, int R, int o, int ql, int qr, int v) {
	int M = L + R >> 1, lc = o << 1, rc = lc | 1;
	if(ql <= L && R <= qr) {
		addv[o] += v;
		if(addv[o]) sumv[o] = A[R] - A[L-1];
		else if(L == R) sumv[o] = 0;
		else sumv[o] = sumv[lc] + sumv[rc];
		return ;
	}
	if(ql <= M) update(L, M, lc, ql, qr, v);
	if(qr > M) update(M+1, R, rc, ql, qr, v);
	sumv[o] = addv[o] ? A[R] - A[L-1] : sumv[lc] + sumv[rc];
	return ;
}

int main() {
	int kase = 0;
	n = read();
	cnt = 0;
	for(int i = 1; i <= n; i++) {
		int l = read(), r = read(), h = read();
		rs[i] = Rec(l, 0, r, h);
		num[++cnt] = l; num[++cnt] = r;
	}
	sort(num + 1, num + cnt + 1);
	cnt = unique(num + 1, num + cnt + 1) - num - 1;
	int tcnt = cnt;
	for(int i = 1; i < cnt; i++) A[i] = num[i+1] - num[1];
	for(int i = 1; i <= n; i++) {
		rsi[i].x1 = lower_bound(num + 1, num + cnt + 1, rs[i].x1) - num;
		rsi[i].x2 = lower_bound(num + 1, num + cnt + 1, rs[i].x2) - num - 1;
	}
	cnt = 0;
	for(int i = 1; i <= n; i++)
		num[++cnt] = rs[i].y1, num[++cnt] = rs[i].y2;
	sort(num + 1, num + cnt + 1);
	cnt = unique(num + 1, num + cnt + 1) - num - 1;
	for(int i = 1; i < cnt; i++) B[i] = num[i+1] - num[i];
	ca = cd = 0;
	for(int i = 1; i <= n; i++) {
		rsi[i].y1 = lower_bound(num + 1, num + cnt + 1, rs[i].y1) - num;
		rsi[i].y2 = lower_bound(num + 1, num + cnt + 1, rs[i].y2) - num;
		ad[++ca] = Line(rsi[i].x1, rsi[i].x2, rsi[i].y1);
		de[++cd] = Line(rsi[i].x1, rsi[i].x2, rsi[i].y2);
	}

	sort(ad + 1, ad + ca + 1);
	sort(de + 1, de + cd + 1);
	LL ans = 0;
	int ka = 1, kd = 1;
	build(1, tcnt, 1);
	for(int i = 1; i <= cnt; i++) {
		while(ka <= ca && ad[ka].x == i)
			update(1, tcnt, 1, ad[ka].l, ad[ka].r, 1), ka++;
		while(kd <= cd && de[kd].x == i)
			update(1, tcnt, 1, de[kd].l, de[kd].r, -1), kd++;
		if(i < cnt) ans += sumv[1] * B[i];
	}
	printf("%lld\n", ans);

	return 0;
}