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            posts - 74,  comments - 33,  trackbacks - 0
            City Horizon
            Time Limit: 2000MS Memory Limit: 65536K
            Total Submissions: 4976 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(shí)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) 評(píng)論(0)  編輯 收藏 引用

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