City Horizon
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 16206 |
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Accepted: 4414 |
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. 和poj1151相同，求解矩形的并的面积http://blog.csdn.net/winddreams/article/details/38495093 按x轴有左向右扫描，对y值进行离散化，为了将每个段连接起来，所以左右子树应该是[l,mid],[mid,r]，每段的差是1，不是0，这样可以防止，忽略掉mid+1到mid这一段，然后判断，只有lazy大于0时，这一个节点的所有长度都加进去，不然节点大小 = 左右子树的和
#include
#include
#include
#include
using namespace std; #define maxn 81000 #define LL __int64 struct node1 { LL x, y1 , y2 ; LL flag ; } p[maxn<<2]; struct node2 { LL l , r ; LL sum ; } cl[maxn<<2]; LL lazy[maxn<<2]; LL s[maxn<<2] ; bool cmp(node1 a,node1 b) { return a.x < b.x ; } void push_up(int rt) { if( lazy[rt] ) cl[rt].sum = cl[rt].r - cl[rt].l ; else cl[rt].sum = cl[rt<<1].sum + cl[rt<<1|1].sum ; } void creat(int rt,int l,int r) { cl[rt].l = s[l] ; cl[rt].r = s[r] ; if( r - l > 1 ) { creat(rt<<1,l,(l+r)/2); creat(rt<<1|1,(l+r)/2,r); } else cl[rt].sum = 0 ; return ; } void update(int rt,LL l,LL r,int flag) { if( l == cl[rt].l && r == cl[rt].r ) { lazy[rt] += flag ; push_up(rt); return ; } else { if( cl[rt<<1].r > l ) { LL k = cl[rt<<1].r < r ? cl[rt<<1].r : r ; update(rt<<1,l,k,flag); } if( cl[rt<<1|1].l < r ) { LL k = cl[rt<<1|1].l > l ? cl[rt<<1|1].l : l ; update(rt<<1|1,k,r,flag); } push_up(rt); } } int main() { LL i , n , x1 , x2 , h , low , ans ; while(scanf("%I64d", &n)!=EOF) { for(i = 0 ; i < n ; i++) { scanf("%I64d %I64d %I64d", &x1, &x2, &h); p[i].x = x1 ; p[i].y1 = 0 ; p[i].y2 = h ; p[i].flag = 1 ; p[i+n].x = x2 ; p[i+n].y1 = 0 ; p[i+n].y2 = h ; p[i+n].flag = -1 ; s[i+1] = h ; } s[n+1] = 0 ; sort(s+1,s+(n+2)); sort(p,p+2*n,cmp); creat(1,1,n+1); update(1,p[0].y1,p[0].y2,p[0].flag); ans = 0 ; for(i = 1 ; i < 2*n ; i++) { ans += ( p[i].x - p[i-1].x )*cl[1].sum; update(1,p[i].y1,p[i].y2,p[i].flag); } printf("%I64d\n", ans); } return 0; }