1645: [Usaco2007 Open]City Horizon 城市地平线
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 315 Solved: 157
[Submit][Status]
Description
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.
N个矩形块，交求面积并.
Input
* 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
Output
* Line 1: The total area, in square units, of the silhouettes formed by all N buildings
Sample Input
4
2 5 1
9 10 4
6 8 2
4 6 3
Sample Output
16
OUTPUT DETAILS:
The first building overlaps with the fourth building for an area of 1
square unit, so the total area is just 3*1 + 1*4 + 2*2 + 2*3 - 1 = 16.
HINT
Source
题解：
这题的思路比较巧妙。
全部的图形被分成了2*n-1个矩形，所以只要用线段树维护每一个矩形的高即可，取max
代码：（copy）
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<algorithm>
6 #define ll long long
7 #define inf 10000000000
8 using namespace std;
9 inline ll read()
10 {
11 int x=0,f=1;char ch=getchar();
12 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
13 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
14 return x*f;
15 }
16 int n;
17 int x[40005],y[40005],val[40005],disc[80005];
18 struct seg{int l,r,mx,tag;}t[320005];
19 int find(int x)
20 {
21 int l=1,r=2*n;
22 while(l<=r)
23 {
24 int mid=(l+r)>>1;
25 if(disc[mid]<x)l=mid+1;
26 else if(disc[mid]==x)return mid;
27 else r=mid-1;
28 }
29 }
30 void pushdown(int k)
31 {
32 if(t[k].l==t[k].r)return;
33 int tag=t[k].tag;t[k].tag=0;
34 if(tag)
35 {
36 t[k<<1].tag=max(t[k<<1].tag,tag);
37 t[k<<1|1].tag=max(t[k<<1|1].tag,tag);
38 t[k<<1].mx=max(t[k<<1].mx,tag);
39 t[k<<1|1].mx=max(t[k<<1|1].mx,tag);
40 }
41 }
42 void build(int k,int l,int r)
43 {
44 t[k].l=l;t[k].r=r;
45 if(l==r)return;
46 int mid=(l+r)>>1;
47 build(k<<1,l,mid);build(k<<1|1,mid+1,r);
48 }
49 void update(int k,int x,int y,int val)
50 {
51 pushdown(k);
52 int l=t[k].l,r=t[k].r;
53 if(l==x&&y==r)
54 {
55 t[k].tag=val;t[k].mx=max(t[k].mx,val);
56 return;
57 }
58 int mid=(l+r)>>1;
59 if(y<=mid)update(k<<1,x,y,val);
60 else if(x>mid)update(k<<1|1,x,y,val);
61 else
62 {
63 update(k<<1,x,mid,val);update(k<<1|1,mid+1,y,val);
64 }
65 }
66 int query(int k,int x)
67 {
68 pushdown(k);
69 int l=t[k].l,r=t[k].r;
70 if(l==r)return t[k].mx;
71 int mid=(l+r)>>1;
72 if(x<=mid)return query(k<<1,x);
73 else return query(k<<1|1,x);
74 }
75 int main()
76 {
77 n=read();build(1,1,n<<1);
78 for(int i=1;i<=n;i++)
79 {
80 x[i]=read(),y[i]=read(),val[i]=read();
81 disc[(i<<1)-1]=x[i];disc[i<<1]=y[i];
82 }
83 sort(disc+1,disc+(n<<1)+1);
84 for(int i=1;i<=n;i++)
85 x[i]=find(x[i]),y[i]=find(y[i]);
86 for(int i=1;i<=n;i++)
87 {
88 update(1,x[i],y[i]-1,val[i]);
89 }
90 ll ans=0;
91 for(int i=1;i<2*n;i++)
92 {
93 ans+=(ll)query(1,i)*(disc[i+1]-disc[i]);
94 }
95 printf("%lld",ans);
96 return 0;
97 }