【POJ 3277 --- City Horizon】离散化+线段树

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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

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

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.

Source

解题思路

线段树+离散化因为它们的宽度太大，无法存储所以先离散，然后再根据按从低到高的顺序更新线段树。

AC代码1：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)#define endl '\n'#define lson root<<1#define rson root<<1|1typedef long long ll;const int MAXN = 4e4+5;ll sum[MAXN<<3];int add[MAXN<<3],sl[MAXN<<3],sr[MAXN<<3];vector
       
         v,vec;struct Node{    int a,b,h;};Node node[MAXN];bool compare(Node a,Node b){    return a.h
        
         >1; build(lson,l,mid); build(rson,mid,r);}void update(int root,int l,int r,int L,int R,int num){ if(l>=L && r<=R) { sum[root]=(ll)num*(sr[root]-sl[root]); add[root]=num; return; } if(l+1==r) return; push_down(root); int mid=(l+r)>>1; if(L<=mid) update(lson,l,mid,L,R,num); if(R>=mid) update(rson,mid,r,L,R,num); push_up(root);}int main(){ SIS; int n; cin >> n; for(int i=0;i
         
          > node[i].a >> node[i].b >> node[i].h; v.push_back(node[i].a); v.push_back(node[i].b); } sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end()); int len=v.size(); build(1,1,len); sort(node,node+n,compare); for(int i=0;i

AC代码2：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)#define endl '\n'#define lson root<<1#define rson root<<1|1typedef long long ll;const int MAXN = 4e4+5;ll sum[MAXN<<3];int add[MAXN<<3],sl[MAXN<<3],sr[MAXN<<3];vector
       
         v,vec;struct Node{    int a,b,h;};Node node[MAXN];bool compare(Node a,Node b){    return a.h
        
         >1; build(lson,l,mid); build(rson,mid,r);}void update(int root,int l,int r,int L,int R,int num){ if(L<=sl[root] && R>=sr[root]) { sum[root]=(ll)num*(sr[root]-sl[root]); add[root]=num; return; } if(l+1==r) return; push_down(root); int mid=(l+r)>>1; if(L<=sr[lson]) update(lson,l,mid,L,R,num); if(R>=sl[rson]) update(rson,mid,r,L,R,num); push_up(root);}int main(){ SIS; int n; cin >> n; for(int i=0;i
         
          > node[i].a >> node[i].b >> node[i].h; v.push_back(node[i].a); v.push_back(node[i].b); } sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end()); int len=v.size(); build(1,1,len); sort(node,node+n,compare); for(int i=0;i

转载地址：http://qiyof.baihongyu.com/

你可能感兴趣的文章