bzoj 1645: [Usaco2007 Open]City Horizon 城市地平线（线段树扫描线）

1645: [Usaco2007 Open]City Horizon 城市地平线

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 732 Solved: 333
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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 A_i through B_i along the horizon (1 <= A_i < B_i <= 1,000,000,000) and has height H_i (1 <= H_i <= 1,000,000,000). Determine the area, in square units, of the aggregate silhouette formed by all N buildings.

N个矩形块，交求面积并.

Input

* Line 1: A single integer: N

* Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: A_i, B_i, and H_i

Output

* Line 1: The total area, in square units, of the silhouettes formed by all N buildings

Sample Input

2 5 1

9 10 4

6 8 2

4 6 3

Sample Output

线段树扫描线模板题

求矩形面积和要比求周长和容易点

#include<stdio.h>
#include<algorithm>
using namespace std;
#define LL long long
typedef struct Tree
{int cnt;int s1;
}Tree;
Tree tre[322233];
typedef struct Line
{int x, y;int flag;bool operator < (const Line &b) const{if(x<b.x || x==b.x && flag<b.flag)return 1;return 0;}
}Line;
Line s[80005];
int pos[80005];
void Atonce(int l, int r, int x)
{if(tre[x].cnt>=1)tre[x].s1 = pos[r]-pos[l-1];else{if(l!=r)tre[x].s1 = tre[x*2].s1+tre[x*2+1].s1;elsetre[x].s1 = 0;}
}
void Update(int l, int r, int x, int h, int val)
{int m;if(r<=h){tre[x].cnt += val;Atonce(l, r, x);return;}m = (l+r)/2;Update(l, m, x*2, h, val);if(h>=m+1)Update(m+1, r, x*2+1, h, val);Atonce(l, r, x);
}
int main(void)
{LL ans;int n, i, x1, x2, cnt, h;scanf("%d", &n);cnt = 0;for(i=1;i<=n;i++){scanf("%d%d%d", &x1, &x2, &h);s[++cnt].x = x1, s[cnt].y = h, s[cnt].flag = 1;s[++cnt].x = x2, s[cnt].y = h, s[cnt].flag = -1;pos[i] = h;}sort(pos+1, pos+n+1);n = unique(pos+1, pos+n+1)-(pos+1);sort(s+1, s+cnt+1);ans = 0;for(i=1;i<=cnt;i++){h = lower_bound(pos+1, pos+n+1, s[i].y)-pos;Update(1, n, 1, h, s[i].flag);ans += (LL)tre[1].s1*(s[i+1].x-s[i].x);}printf("%lld\n", ans);return 0;
}