UVA1232,LA4108,Skyline，线段树

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.

Input

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. The i-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.

Output

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.

Sample Input

1 
3 
5 11 3 
1 10 1 
3 13 2 
0

Sample Output

14

题意：建筑物在多长的部分是最高的成为该建筑物的覆盖度。求所有建筑物的覆盖度之和。

思路：[l,r]的建筑物可以用[l+1,r]来表示。为什么呢？自己在草稿纸上画画图~~

           线段树维护set,minh,maxh。即区间的最小值，区间的最大值。。

           当要进行区间置值，只要区间的最大值小于等于懒惰标记。则可以直接置值
           当要进行区间置值，如果区间的最小值大于懒惰标记。可忽略、

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 100080
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
//线段树要维护区间最高的建筑，还要有懒惰标记。
//当要进行区间置值，只要区间的最大值小于等于懒惰标记。则可以直接置值
//当要进行区间置值，如果区间的最小值大于懒惰标记。可忽略、
struct ST
{
	int l,r,set,maxh,minh;
}st[maxn<<2];
int ans;
inline int max(int a,int b)
{
	return a>b?a:b;
}

inline int min(int a,int b)
{
	return a>b?b:a;
}

void PushDown(int id)
{
	if(st[id].set != -1)
	{
		st[id<<1].set = st[id<<1|1].set = st[id].set;
		st[id<<1].maxh = st[id<<1|1].maxh = st[id].set;
		st[id<<1].minh = st[id<<1|1].minh = st[id].set;
		st[id].set = -1;
	}
}

void PushUp(int id)
{
	st[id].maxh = max(st[id<<1].maxh,st[id<<1|1].maxh);
	st[id].minh = min(st[id<<1].minh,st[id<<1|1].minh);
}

void buildtree(int id,int l,int r)
{
	st[id].l = l,st[id].r = r;
	st[id].set = st[id].maxh = st[id].minh = 0;
	if(l == r)	return;
	int mid = (l+r) >> 1;
	buildtree(lson);
	buildtree(rson);
}

void Update(int id,int l,int r,int key)
{
	if(st[id].minh > key)	return;
	if(st[id].l == l && st[id].r == r)
	{
		if(st[id].maxh <= key)
		{
			st[id].set = st[id].maxh = st[id].minh = key;
			ans += r - l + 1;
			return;
		}
		if(l == r)	return;
	}
	PushDown(id);
	if(st[id<<1].r >= r)
		Update(id<<1,l,r,key);
	else if(st[id<<1|1].l <= l)
		Update(id<<1|1,l,r,key);
	else 
	{
			Update(id<<1,l,st[id<<1].r,key);
			Update(id<<1|1,st[id<<1|1].l,r,key);
	}
	PushUp(id);
}

int main()
{
	//freopen("in.txt","r",stdin);
	int t;
	while(scanf("%d",&t)!=EOF && t)
	{
		while(t--)
		{
			int n;
			scanf("%d",&n);
			ans = 0;
			buildtree(1,1,100010);
			for(int i = 1;i <= n;i++)
			{
					int l,r,key;	scanf("%d%d%d",&l,&r,&key);
					Update(1,l+1,r,key);
			}
			printf("%d\n",ans);
		}
	}
	return 0;
}