uva1232 - 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

  按顺序建房子，给出房子的左右坐标和高度，每个房子的覆盖度为它建的时候覆盖区间最高（相等也算）的区间长度。求总覆盖度。

  用懒惰标记，设minv[o]，maxv[o]，setv[o]分别为经过以o节点为根的树内的操作o区间范围内的最小值，最大值和置值。查询时候就算[L,R]在[ql,qr]区间内也必须要height>=maxv[o]或者L==R的时候才能把区间置值然后return，因为如果全部覆盖了才能set，否则就要把set往下推一层，然后继续递归，维护节点信息。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<stack>
#define INF 0x3f3f3f3f
#define MAXN 100010
#define MAXM 35
#define MAXNODE 4*MAXN
#define eps 1e-9
#define pi 4*atan(1.0)
#define pii pair<int,int>
using namespace std;
int T,N,ql,qr,height,ans,minv[MAXNODE],maxv[MAXNODE],setv[MAXNODE],l[MAXN],r[MAXN],h[MAXN];
void maintain(int o,int L,int R){
    int lc=(o<<1),rc=(o<<1|1);
    if(L<R){
        minv[o]=min(minv[lc],minv[rc]);
        maxv[o]=max(maxv[lc],maxv[rc]);
    }
    if(setv[o]!=-1) minv[o]=maxv[o]=setv[o];
}
void pushdown(int o){
    if(setv[o]!=-1){
        int lc=(o<<1),rc=(o<<1|1);
        setv[lc]=setv[rc]=setv[o];
        setv[o]=-1;
    }
}
void update(int o,int L,int R){
    if(height<minv[o]) return;
    if(ql<=L&&qr>=R){
        if(height>=maxv[o]){
            setv[o]=maxv[o]=minv[o]=height;
            ans+=R-L+1;
            return;
        }
        if(L==R) return;
    }
    pushdown(o);
    int mid=(L+R)/2,lc=(o<<1),rc=(o<<1|1);
    if(ql<=mid) update(lc,L,mid);
    else maintain(lc,L,mid);
    if(qr>mid) update(rc,mid+1,R);
    else maintain(rc,mid+1,R);
    maintain(o,L,R);
}
int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d",&T)!=EOF,T){
        while(T--){
            memset(setv,-1,sizeof(setv));
            memset(minv,0,sizeof(minv));
            memset(maxv,0,sizeof(maxv));
            setv[1]=0;
            scanf("%d",&N);
            int MINL=INF,MAXR=-1;
            for(int i=0;i<N;i++){
                scanf("%d%d%d",&l[i],&r[i],&h[i]);
                MINL=min(MINL,l[i]);
                MAXR=max(MAXR,r[i]);
            }
            ans=0;
            for(int i=0;i<N;i++){
                ql=l[i]+1;
                qr=r[i];
                height=h[i];
                update(1,MINL,MAXR);
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}