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
4
2 5 1
9 10 4
6 8 2
4 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.
Source
题目的意思是说：平面上有[1,40000]个建筑，每个建筑有一个区间[Ai,Bi]表示它的跨度，Hi表示其高度。要求这n个建筑的平面覆盖面积。
由于Ai，Bi∈[1,1000000000]，直接将线段[Ai,Bi]插入线段树空间消耗太大，可以将这n条线段的2n个端点离散化到一个数组X[0...2n-1]，然后再将其插入线段树，最后求出面积。
#include <iostream>
using namespace std;
const int MAXN = 40010;
struct segment
{
int left,right,h;
}tree[MAXN*3];
int pos[MAXN][3],x[MAXN*2],len;
int cmp1(const void *a,const void *b){
return *(int *)a-*(int *)b;
}
int cmp2(const void *a,const void *b){
return *((int *)a+2)-*((int *)b+2);
}
int query(int h){
int low=0,high=len-1,mid;
while(low<=high){
mid=(low+high)>>1;
if(x[mid]==h) return mid;
else if(x[mid]>h) high=mid-1;
else low=mid+1;
}
return -1;
}
void create(int l,int r,int index){
tree[index].left=l,tree[index].right=r;
tree[index].h=0;
if(r==l+1) return;
int mid=(l+r)>>1;
create(l,mid,2*index);
create(mid,r,2*index+1);
}
void update(int l,int r,int h,int index){
if(h<tree[index].h) return;
if(l==tree[index].left && r==tree[index].right){
tree[index].h=h;
return;
}
if(tree[index].h>=0 && tree[index].right>tree[index].left+1){
tree[2*index].h=tree[2*index+1].h=tree[index].h;
tree[index].h=-1;
}
int mid=(tree[index].left+tree[index].right)>>1;
if(r<=mid)
update(l,r,h,2*index);
else if(l>=mid)
update(l,r,h,2*index+1);
else{
update(l,mid,h,2*index);
update(mid,r,h,2*index+1);
}
}
long long cal(int index){
if(tree[index].h>=0)
return tree[index].h*(long long)(x[tree[index].right]-x[tree[index].left]);
else
return cal(2*index)+cal(2*index+1);
}
int main(){
int n,i,k,l,r;
scanf("%d",&n);
for(i=len=0;i<n;i++,len+=2){
scanf("%d %d %d",&pos[i][0],&pos[i][1],&pos[i][2]);
x[len]=pos[i][0],x[len+1]=pos[i][1];
}
qsort(x,2*n,sizeof(int),cmp1);
for(i=1,k=0;i<2*n;i++)
if(x[i]!=x[i-1]) x[k++]=x[i-1];
x[k++]=x[i-1],len=k;
qsort(pos,n,3*sizeof(int),cmp2);
create(0,len-1,1);
for(i=0;i<n;i++){
l=query(pos[i][0]),r=query(pos[i][1]);
update(l,r,pos[i][2],1);
}
printf("%I64d\n",cal(1));
return 0;
}