Poj 3277 city horizon

In simple lazy operations, you can push all lazy operations to the leaf node during statistics.

City horizon

Time limit:2000 ms		Memory limit:65536 K
Total submissions:15973		Accepted:4336

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 lineN(1 ≤N≤ 40,000) buildings. BuildingI'S silhouette has a base that spans locationsAIThroughBiAlong the horizon (1 ≤AI<Bi≤ 1,000,000,000) and has heightHi(1 ≤Hi≤ 1,000,000,000). determine the area, in square units, of the aggregate silhouette formed by allNBuildings.

Input

Line 1: A single INTEGER: N 
Lines 2 .. N+ 1: input line I+ 1 describes Building IWith three space-separated integers: AI, Bi, And Hi

Output

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

Sample Input

42 5 19 10 46 8 24 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

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;typedef long long int LL;const LL maxn=204000;LL r[maxn*2],md[maxn*2],val[maxn*2],n,m,vt,zhi[maxn*2],Hash[maxn*2],high[maxn];bool cmpR(int a,int b){   return val[a]<val[b];}int Lisan(int n){    for(int i=0;i<n;i++) r[i]=i;    sort(r,r+n,cmpR);    md[0]=val[r[0]];    val[r[0]]=m=0;    for(int i=1;i<n;i++)    {        if(md[m]!=val[r[i]])            md[++m]=val[r[i]];        val[r[i]]=m;    }    return m;}LL tree[maxn<<2],cover[maxn<<2];void push_down(int l,int r,int rt){    if(cover[rt])    {        tree[rt<<1]=max(tree[rt<<1],tree[rt]);        tree[rt<<1|1]=max(tree[rt<<1|1],tree[rt]);        cover[rt<<1]=cover[rt<<1|1]=1;        cover[rt]=0;    }}void update(int L,int R,LL c,int l,int r,int rt){    if(L<=l&&r<=R)    {        tree[rt]=max(tree[rt],c);        cover[rt]=1;        return ;    }    push_down(l,r,rt);    int m=(l+r)/2;    if(L<=m) update(L,R,c,lson);    if(R>m) update(L,R,c,rson);}LL ans;void over_tree(int l,int r,int rt){    push_down(l,r,rt);    if(l==r)    {        ans+=(Hash[r+1]-Hash[l])*tree[rt];        return ;    }    int m=(l+r)/2;    over_tree(lson); over_tree(rson);}int main(){    while(scanf("%d",&n)!=EOF)    {        vt=0;        memset(tree,0,sizeof(tree));        memset(cover,0,sizeof(cover));        for(int i=0;i<n;i++)        {            LL a,b,c;            scanf("%I64d%I64d%I64d",&a,&b,&c);            val[vt]=zhi[vt]=a; vt++;            val[vt]=zhi[vt]=b; vt++;            high[i]=c;        }        int mx=Lisan(vt);        for(int i=0;i<vt;i++)        {            Hash[val[i]]=zhi[i];        }        for(int i=0;i<n;i++)        {            update(val[i*2],val[i*2+1]-1,high[i],0,mx,1);        }        ans=0;        over_tree(0,mx,1);        printf("%I64d\n",ans);    }    return 0;}