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
題目的意思是說(shuō)：平面上有[1,40000]個(gè)建筑，每個(gè)建筑有一個(gè)區(qū)間[Ai,Bi]表示它的跨度，Hi表示其高度。要求這n個(gè)建筑的平面覆蓋面積。
由于Ai，Bi∈[1,1000000000]，直接將線段[Ai,Bi]插入線段樹(shù)空間消耗太大，可以將這n條線段的2n個(gè)端點(diǎn)離散化到一個(gè)數(shù)組X[0...2n-1]，然后再將其插入線段樹(shù)，最后求出面積。
#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;
}