UVALive - 4108 SKYLINE[线段树]

Posted 2020-08-14 Candy?

UVALive - 4108 SKYLINE Time Limit: 3000MS     64bit IO Format: %lld & %llu Submit Status uDebug Description   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.    <tex2html_verbatim_mark> 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,..., bn <tex2html_verbatim_mark>, 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 <tex2html_verbatim_mark>buildings is the union of the rectangles of the first k <tex2html_verbatim_mark>buildings (see Figure 4). The overlap of a building, bi <tex2html_verbatim_mark>, is defined as the total horizontal length of the parts of bi <tex2html_verbatim_mark>, 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 bi<tex2html_verbatim_mark>which has a height that is less than or equal to hi <tex2html_verbatim_mark>, where hi <tex2html_verbatim_mark>is the height of building bi <tex2html_verbatim_mark>. You may assume that initially the skyline has height zero everywhere.    Input The input consists of a line containing the number c <tex2html_verbatim_mark>of datasets, followed by c <tex2html_verbatim_mark>datasets, followed by a line containing the number ` 0‘.  The first line of each dataset consists of a single positive integer, n <tex2html_verbatim_mark>(0 < n < 100000) <tex2html_verbatim_mark>, which is the number of buildings in the proposal. The following n<tex2html_verbatim_mark>lines of each dataset each contains a description of a single building. The i <tex2html_verbatim_mark>-th line is a description of building bi <tex2html_verbatim_mark>. Each building bi <tex2html_verbatim_mark>is described by three positive integers, separated by spaces, namely, li <tex2html_verbatim_mark>, ri <tex2html_verbatim_mark>and hi <tex2html_verbatim_mark>, where li <tex2html_verbatim_mark>and rj <tex2html_verbatim_mark>(0 < li < ri100000) <tex2html_verbatim_mark>represents the left and right end point of the building and hi represents the height of the building.    <tex2html_verbatim_mark>   Output The output consists of one line for each dataset. The c <tex2html_verbatim_mark>-th line contains one single integer, representing the amount of overlap in the proposal for dataset c<tex2html_verbatim_mark>. 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 <tex2html_verbatim_mark>, b2 <tex2html_verbatim_mark>and b3 <tex2html_verbatim_mark>are 6, 4 and 4 respectively. Figure 4 shows how to compute the overlap of building b3 <tex2html_verbatim_mark>. The grey area represents the skyline formed by b1 <tex2html_verbatim_mark>and b2 <tex2html_verbatim_mark>and the black rectangle represents b3 <tex2html_verbatim_mark>. As shown in the figure, the length of the skyline covered by b3 <tex2html_verbatim_mark>is from position 3 to position 5 and from position 11 to position 13, therefore the overlap of b3 <tex2html_verbatim_mark>is 4.    Sample Input   1 3 5 11 3 1 10 1 3 13 2 0   Sample Output   14

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题意：在线区间[l,r]查询h是最大值的长度，并更新

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很明显区间max

但怎么找这个长度overlap呢？

如果当前区间满足ql<=l&&r<=qr&&h>=t[o].mx 那么就可以被完全覆盖

用lazy维护区间完全覆盖中的最大值，h<t[o].lazy时就不用继续找了(因为h不可能覆盖这个区间了)，有点像个剪枝

否则必须继续找，下传lazy(无需清空自己的标记,剪枝嘛)，合并mx

lazy和mx还有点小细节，比如paint没有用lazy更新mx，其实这不影响，因为先if(h<t[o].lazy) return;了

注意点是区间[i,i+1]

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define m (l+r)/2
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define lc o<<1
#define rc o<<1|1
using namespace std;
typedef long long ll;
const int N=1e5+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}
int T,n,ql,qr,h;
struct node{
    int mx,lazy;//totally cover
}t[N<<2];
inline void pushDown(int o){
    if(t[o].lazy>t[lc].lazy) t[lc].lazy=t[o].lazy;
    if(t[o].lazy>t[rc].lazy) t[rc].lazy=t[o].lazy;
}
int lap=0;
inline void update(int o,int l,int r,int ql,int qr,int h){
    if(h<t[o].lazy) return;
    if(ql<=l&&r<=qr&&h>=t[o].mx){
        t[o].mx=t[o].lazy=h;
        lap+=r-l+1;
    }else{
        pushDown(o);
        if(ql<=m) update(lson,ql,qr,h);
        if(m<qr) update(rson,ql,qr,h);
        t[o].mx=max(t[lc].mx,t[rc].mx);
    }
}

int main(){
    T=read();
    while(T--){
        n=read();
        int ans=0;
        memset(t,0,sizeof(t));
        for(int i=1;i<=n;i++){
            ql=read();qr=read()-1;h=read();
            lap=0;
            update(1,1,1e5,ql,qr,h);
            //printf("lap %d\n",lap);
            ans+=lap;
        }
        printf("%d\n",ans);
    }
}