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POJ 3277 City Horizon
City Horizon
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 18555 | Accepted: 5114 |
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 Ai through Bi along the horizon (1 ≤ Ai < Bi ≤ 1,000,000,000) and has height Hi(1 ≤ Hi ≤ 1,000,000,000). Determine the area, in square units, of the aggregate silhouette formed by all N buildings.
Input
Line 1: A single integer: N
Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: Ai, Bi, and Hi
Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: Ai, Bi, and Hi
Output
Line 1: The total area, in square units, of the silhouettes formed by all N buildings
Sample Input
42 5 19 10 46 8 24 6 3
Sample Output
16
Hint
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.
题目大意：在水平直线上给定一组矩形，这些矩形的底边在同一水平线上，求这些矩形占据的面积。
题目分析：很显然，这些矩形的长和高都不一样，还会有重合的区域。如何求这些矩形的面积呢？画个图就知道了。
本题需要注意：
离散化的手法；
延迟更新的手法：在最后求面积时延迟
long long （强制转换）；
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 7 #define lson l,m,rt<<1 8 #define rson m,r,rt<<1|1 9 10 typedef long long LL;11 12 const int maxn=40005;13 const int maxl=maxn*2;14 int a[maxn],b[maxn],c[maxn];15 int jh[maxl];16 int h[maxl<<2],tag[maxl<<2];17 int N,M;18 19 void UpDate(int L,int R,int d,int l,int r,int rt)20 {21 if(L<=l&&r<=R)22 {23 if(h[rt]<d) h[rt]=d;24 return;25 }26 if(l+1==r) return;27 int m=(l+r)>>1;28 if(L<=m) UpDate(L,R,d,lson);29 if(R>=m) UpDate(L,R,d,rson);30 }31 32 LL Query(int l,int r,int rt,int height)//lazy lazy lazy33 {34 if(height>h[rt]) h[rt]=height;35 if(l+1==r) return (LL)(jh[r]-jh[l])*h[rt];36 int m=(l+r)>>1;37 LL a=Query(lson,h[rt]);38 LL b=Query(rson,h[rt]);39 return a+b;40 }41 42 int bfind(int x)43 {44 int l,r;45 l=0,r=M+1;46 while(l<r)47 {48 int m=(l+r)>>1;49 if(jh[m]==x) return m;50 if(jh[m]<x) l=m;51 else r=m;52 }53 }54 55 int main()56 {57 scanf("%d",&N);58 for(int i=1;i<=N;i++)59 {60 scanf("%d %d %d",&a[i],&b[i],&c[i]);61 jh[2*i-1]=a[i],jh[2*i]=b[i];62 }63 64 sort(jh+1,jh+2*N+1);65 for(int i=1;i<=2*N;i++)66 if(jh[i+1]!=jh[i]) jh[++M]=jh[i];67 for(int i=1;i<=N;i++)68 {69 if(a[i]>b[i]) swap(a[i],b[i]);70 UpDate(bfind(a[i]),bfind(b[i]),c[i],1,M,1);71 }72 int m=(1+M)>>1;73 cout<<Query(1,m,2,h[1])+Query(m,M,3,h[1]);74 return 0;75 }
POJ 3277 City Horizon
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