poj 3277(线段树+离散化)

City Horizon

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 18646 Accepted: 5145

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.

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

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.

//坐标 x离散化  离散后求相邻两点的最大值h 接着对应原始的x坐标求面积 这里需要注意离散化卡map 所以可以利用二分查找  还有由于更新会导致左右两个坐标都被覆盖，所以离散化时可以 1 3 5 7。。。这样离散 每次求2 4 6 8的值即可 

#include <algorithm>#include <iostream>#include <map>#include <set>#include <cstring>#include <cstdio>using namespace std;#define ll long longconst int N=1e6+5;int n;int l[N],r[N],h[N];set<int>s;set<int>::iterator it;int Map[2*N];ll ans;void init(){    s.clear();    memset(Map,0,sizeof(Map));    memset(l,0,sizeof(l));    memset(r,0,sizeof(r));    memset(h,0,sizeof(h));}struct Node{    int l,r;    int Ma;}seg[4*N];void PushDown(int rt){    int t=seg[rt].Ma;    if(t>seg[2*rt].Ma)seg[2*rt].Ma=t;    if(t>seg[2*rt+1].Ma)seg[2*rt+1].Ma=t;}void Build(int rt,int l,int r){    seg[rt].l=l,seg[rt].r=r;    seg[rt].Ma=0;    if(l!=r)    {        Build(2*rt,l,(l+r)/2);        Build(2*rt+1,(l+r)/2+1,r);    }}void Update(int rt,int l,int r,int val){    if(seg[rt].l==l&&seg[rt].r==r)    {        if(val>seg[rt].Ma)seg[rt].Ma=val;        return;    }    PushDown(rt);    int mid=(seg[rt].l+seg[rt].r)/2;    if(r<=mid)Update(2*rt,l,r,val);    else if(l>mid)Update(2*rt+1,l,r,val);    else    {        Update(2*rt,l,mid,val);        Update(2*rt+1,mid+1,r,val);    }}void solve(int rt,int l,int r){    PushDown(rt);    if(l==r)    {        if(r%2==0)        {            int mi=seg[rt].r;            ans+=(ll)(seg[rt].Ma)*(Map[mi+1]-Map[mi-1]);        }        return;    }    else    {        int mid=(l+r)/2;        solve(2*rt,l,mid);        solve(2*rt+1,mid+1,r);    }}int main(){    while(~scanf("%d",&n))    {        init();        for(int i=1;i<=n;i++)        {            scanf("%d%d%d",l+i,r+i,h+i);            s.insert(l[i]),s.insert(r[i]);        }        int cnt=1;        for(it=s.begin();it!=s.end();it++) //二分查找时由于是1 3 5  所以需要把 2 4 6补上        {            int t=*it;          //  cout<<t<<endl;            Map[cnt]=t;          //  cout<<cnt<<endl;            Map[cnt+1]=t;            cnt+=2;        }        cnt-=2;        Build(1,1,cnt);        for(int i=1;i<=n;i++)        {            int L=l[i],R=r[i];            L=lower_bound(Map+1,Map+cnt+1,L)-Map;            R=lower_bound(Map+1,Map+cnt+1,R)-Map;            Update(1,L,R,h[i]);        }        ans=0;        solve(1,1,cnt);        printf("%I64d\n",ans);    }    return 0;}