[LA4108]SKYLINE

[LA4108]SKYLINE

试题描述

The skyline of Singapore as viewed from the Marina Promenade (shown on the left) is one of the iconic scenes of Singapore. Country X would also like to create an iconic skyline, and it has put up a call for proposals. Each submitted proposal is a description of a proposed skyline and one of the metrics that country X will use to evaluate a proposed skyline is the amount of overlap in the proposed sky-line.

As the assistant to the chair of the skyline evaluation committee, you have been tasked with determining the amount of overlap in each proposal. Each proposal is a sequence of buildings, b₁,b₂,..., b_n , where a building is specified by its left and right endpoint and its height. The buildings are specified in back to front order, in other words a building which appears later in the sequence appears in front of a building which appears earlier in the sequence.

The skyline formed by the first k buildings is the union of the rectangles of the first k buildings (see Figure 4). The overlap of a building, b_i , is defined as the total horizontal length of the parts of b_i , whose height is greater than or equal to the skyline behind it. This is equivalent to the total horizontal length of parts of the skyline behind b_i which has a height that is less than or equal to h_i , where h_i is the height of building b_i . You may assume that initially the skyline has height zero everywhere.

输入

The input consists of a line containing the number c of datasets, followed by c datasets, followed by a line containing the number `0'.

The first line of each dataset consists of a single positive integer, n (0 < n < 100000) , which is the number of buildings in the proposal. The following n lines of each dataset each contains a description of a single building. Thei -th line is a description of building b_i . Each building b_i is described by three positive integers, separated by spaces, namely, l_i , r_i and h_i , where l_i and r_j (0 < l_i < r_i100000) represents the left and right end point of the building and h_i represents the height of the building.

输出

The output consists of one line for each dataset. The c -th line contains one single integer, representing the amount of overlap in the proposal for dataset c . You may assume that the amount of overlap for each dataset is at most 2000000.

Note: In this test case, the overlap of building b₁ , b₂ and b₃ are 6, 4 and 4 respectively. Figure 4 shows how to compute the overlap of building b₃ . The grey area represents the skyline formed by b₁ and b₂ and the black rectangle represents b₃ . As shown in the figure, the length of the skyline covered by b₃ is from position 3 to position 5 and from position 11 to position 13, therefore the overlap of b₃ is 4.

输入示例

输出示例

数据规模及约定

见“输入”

题解

为尊重原题面我就不翻译了。线段树打懒标记，记录一下区间最大最小值，小于最小值直接剪枝，大于等于最大值直接打懒标记并记录答案，否则下传标记接着递归处理。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
 
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
    if(Head == Tail) {
        int l = fread(buffer, 1, BufferSize, stdin);
        Tail = (Head = buffer) + l;
    }
    return *Head++;
}
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 100010
int n, q, ans;
 
int maxv[maxn<<2], minv[maxn<<2], setv[maxn<<2];
void build(int L, int R, int o) {
	maxv[o] = minv[o] = 0; setv[o] = -1;
	if(L == R) return ;
	int M = L + R >> 1, lc = o << 1, rc = lc | 1;
	build(L, M, lc); build(M+1, R, rc);
	return ;
}
void pushdown(int o) {
	int lc = o << 1, rc = lc | 1;
	if(setv[o] >= 0) {
		maxv[o] = minv[o] = setv[o];
		setv[lc] = setv[rc] = setv[o];
		setv[o] = -1;
	}
	return ;
}
int ql, qr;
void update(int L, int R, int o, int h) {
	pushdown(o);
	if(minv[o] > h) return ;
	if(ql <= L && R <= qr && maxv[o] <= h) {
		maxv[o] = minv[o] = setv[o] = h;
		ans += (R - L + 1);
		return ;
	}
	int M = L + R >> 1, lc = o << 1, rc = lc | 1;
	if(ql <= M) update(L, M, lc, h);
	if(qr > M) update(M+1, R, rc, h);
	maxv[o] = max(maxv[lc], maxv[rc]);
	minv[o] = min(minv[lc], minv[rc]);
	return ;
}
 
int main() {
	int T = read();
	while(T--) {
		q = read(); ans = 0;
		n = maxn - 10;
		build(1, n, 1);
		while(q--) {
			ql = read(); qr = read() - 1; int h = read();
			update(1, n, 1, h);
		}
		printf("%d\n", ans);
	}
 
	return 0;
}