xtu数据结构 H. City Horizon

H. City Horizon

Time Limit: 2000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld Java class name: Main

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

Sample Output

16

解题：看了解题报告的，还在领悟离散化是什么灰机，不过这题目线段树部分其实很容易的，只是加上了离散化这一操作，Eggache啊！

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cstdlib>

 #include <vector>

 #include <climits>

 #include <algorithm>

 #include <cmath>

 #define LL long long

 #define INF 0x3f3f3f

 using namespace std;

 const int maxn = ;

 struct Building{

     LL a,b,h;

 }bb[maxn];

 struct node{

     int lt,rt,h;

 }tree[maxn<<];

 LL p[maxn<<],n,ans;

 int cnt;

 bool cmp(const Building &x,const Building &y){

     return x.h < y.h;

 }

 void build(int lt,int rt,int v){

     tree[v].lt = lt;

     tree[v].rt = rt;

     tree[v].h = ;

     if(rt-lt == ) return;

     int mid = (lt+rt)>>;

     build(lt,mid,v<<);

     build(mid,rt,v<<|);

 }

 void update(int lt,int rt,int val,int v){

     if(tree[v].lt == lt && tree[v].rt == rt){

         tree[v].h = val;

         return;

     }

     if(tree[v].h > ){

         tree[v<<].h = tree[v<<|].h = tree[v].h;

         tree[v].h = ;

     }

     int mid = (tree[v].lt+tree[v].rt)>>;

     if(rt <= mid) update(lt,rt,val,v<<);

     else if(lt >= mid) update(lt,rt,val,v<<|);

     else{

         update(lt,mid,val,v<<);

         update(mid,rt,val,v<<|);

     }

 }

 void query(int v){

     if(tree[v].h > ){

         ans += (LL)tree[v].h*(p[tree[v].rt-] - p[tree[v].lt-]);

         return;

     }

     if(tree[v].rt - tree[v].lt == ) return;

     query(v<<);

     query(v<<|);

 }

 int bsearch(int lt,int rt,int val){

     while(lt <= rt){

         int mid = (lt+rt)>>;

         if(p[mid] == val)

             return mid+;

         else if(val < p[mid])

          rt = mid-;

         else lt = mid+;

     }

     return ;

 }

 int main(){

     int i,j,r;

     scanf("%lld",&n);

     for(i = ; i < n; i++){

         scanf("%d%d%d",&bb[i].a,&bb[i].b,&bb[i].h);

         p[cnt++] = bb[i].a;

         p[cnt++] = bb[i].b;

     }

     sort(p,p+cnt);//排序是为了使用二分查找。

     sort(bb,bb+n,cmp);

     build(,n<<,);

     r = (n<<)-;

     for(i = ; i < n; i++){

         int lt = bsearch(,cnt-,bb[i].a);

         int rt = bsearch(,cnt-,bb[i].b);

         update(lt,rt,bb[i].h,);

     }

     ans = ;

     query();

     printf("%lld",ans);

     return ;

 }