City Horizon
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Time Limit: 2000MS |
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Memory Limit: 65536K |
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Total Submissions: 4976 |
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Accepted: 1195 |
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
USACO 2007 Open Silver
這道題目屬于區(qū)間覆蓋，和count colour那一道題目屬于同一類型，我就偷懶了，就直接把那道題的代碼直接copy過來，
沒想到在改的時候，多刪了一句話導(dǎo)致TLE了20+次，很不happy
主要思路代碼如下：
void?Build(int?now,int?l,int?r)
{
????ST[now].l=l,ST[now].r=r,ST[now].h=0,ST[now].mark=true;
????if(l+1>=r)return;
????int?mid=(l+r)>>1;
????Build(2*now,l,mid);
????Build(2*now+1,mid,r);????
????return?;
}
void?insert(int?now,int?l,int?r,int?h){????
????if(ST[now].mark&&ST[now].h>h)return;
????if(ST[now].mark&&ST[now].l==l&&ST[now].r==r){
????????ST[now].h=h;return?;
????}????
????if(ST[now].mark&&ST[now].l+1<ST[now].r){
????????ST[2*now].h=ST[2*now+1].h=ST[now].h;
????????ST[2*now].mark=ST[2*now+1].mark=true;
????????ST[now].mark=false;
????}
????int?mid=(ST[now].l+ST[now].r)>>1;
????if(l>=mid)insert(2*now+1,l,r,h);
????else?if(r<=mid)insert(2*now,l,r,h);
????else?{????
????????insert(2*now,l,mid,h);
????????insert(2*now+1,mid,r,h);
????}????
????return?;
}
posted on 2009-04-08 14:57
KNIGHT 閱讀(110)
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