zhber 有好多做过的题没写下来，如果我还能记得就补吧

## 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

4
2 5 1
9 10 4
6 8 2
4 6 3

## Sample Output

16

OUTPUT DETAILS:

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.

#include<cstdio>
#include<algorithm>
#define LL long long
#define N 50010
#define mod 1000007
using namespace std;
struct trees{
int l,r,mx;
}tree[8*N];
int l,r,mx;
}a[N];
int n,treesize;
LL ans;
int num[2*N];
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline void pushdown(int now)
{
if (tree[now].l==tree[now].r)return;
int mx=tree[now].mx;tree[now].mx=0;
if (mx)
{
tree[now<<1].mx=mx;
tree[now<<1|1].mx=mx;
}
}
inline void buildtree(int now,int l,int r)
{
tree[now].l=l;
tree[now].r=r;
if (l==r)return;
int mid=(l+r)>>1;
buildtree(now<<1,l,mid);
buildtree(now<<1|1,mid+1,r);
}
inline void change(int now,int x,int y,int mx)
{
pushdown(now);
int l=tree[now].l,r=tree[now].r;
if (l==x&&r==y)
{
tree[now].mx=mx;
return;
}
int mid=(l+r)>>1;
if (y<=mid) change(now<<1,x,y,mx);
else if(x>mid)change(now<<1|1,x,y,mx);
else
{
change(now<<1,x,mid,mx);
change(now<<1|1,mid+1,y,mx);
}
}
inline void dfs(int now)
{
int l=tree[now].l,r=tree[now].r;
if (tree[now].mx)
{
ans+=(LL)tree[now].mx*(num[r+1]-num[l]);
return;
}
if (l==r)return;
dfs(now<<1);
dfs(now<<1|1);
}
//----------------------------------离散
struct hashing{
int num,next,rnk;
}hash[mod];
int ha[2*N],len,cnt,rating;
inline void insert(int u,int v,int w)
{
hash[++cnt].num=v;
hash[cnt].rnk=w;
}
inline int find(int x)
{
int s=x%mod;
if (hash[i].num==x)return hash[i].rnk;
}
{return a.mx<b.mx||a.mx==b.mx&&a.l<b.l||a.mx==b.mx&&a.l==b.l&&a.r<b.r;}
//----------------------------------end
int main()
{
for (int i=1;i<=n;i++)
{
ha[++len]=a[i].l;
ha[++len]=a[i].r;
}
sort(ha+1,ha+len+1);
for (int i=1;i<=len;i++)
if (ha[i]!=ha[i-1])
{
num[++rating]=ha[i];
insert(ha[i]%mod,ha[i],rating);
}
for (int i=1;i<=n;i++)
{
a[i].l=find(a[i].l);
a[i].r=find(a[i].r);
}
sort(a+1,a+n+1,cmp);
buildtree(1,1,rating-1);
for (int i=1;i<=n;i++)
change(1,a[i].l,a[i].r-1,a[i].mx);
dfs(1);
printf("%lld",ans);
}

posted on 2014-07-30 22:54  zhber  阅读(182)  评论(0编辑  收藏  举报