题目链接：
题意：就是给出狠点建筑的坐标和高度，然后给出很多询问，求在这个坐标右下角的第k矮的建筑。。
思路：太弱了我，这个题目从上个星期天就开始看，但是一直不会，所以只能看别人思路，因为那个k小于10，所以左右节点只取前十就可以了，但是我觉得万一不记录完全万一发生丢失怎么办，后来一想sb了，如果左右节点都取前10的话，那么根节点得到的20个值，在排序必定取到了前10啊。。。言归正传，这道题因为数据范围太大了，前对建筑和询问一起排序，按x从小到大，再按y从小到大，在按建筑优先的原则排序。所以先对坐标y进行离散化，然后建树，然后开始插入，然后对节点进行维护（只需要维护前10个值就可以了），然后这个问题就解决了。。。我觉得和cf的一个题很像。。。
ps:我这个题敲了几遍，觉得收货很大。。还有就是这题数据太弱了，直接暴力也可以过，并且跑的比线段树的快多了。。。
题目：
K-short Problem
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 366 Accepted Submission(s): 98
Problem Description
In the view of a satellite, you can see lots of tall buildings in the city, and we assume that the city‘s border is flat. Now you have observed n tall buildings in the city and recorded their position and height. Due to some mysterious reasons, you need to
answer a series of queries. Each query will give you a position(x, y) and k, then you have to find out the k-th shortest building in set{(x,y)|x≤X,y≤Y}.
Input
Multiple test cases.For each test case, the first line will contain two integers n and m(0<n≤30000,0≤m≤30000),
which represents the amount of buildings and amount of queries. Then n lines follow, contains three integers x,y,h(?1E9≤x,y≤1E9,0≤h≤1E9) indicate
the building‘s position and height in each line. Then there are m lines, each with three integers x,y,k(?1E9≤x,y≤1E9,1≤k≤10) used
for query.
Output
For each test case, if the k-th shortest building exists, output its height in 1 line, otherwise print "-1" (without quotes).
Sample Input
5 2 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 0 0 1 6 3 2
Sample Output
-1 2
Source
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代码:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=60000+10;
struct Node
{
int x,y,id,flag,h;
void init()
{
scanf("%d%d%d",&x,&y,&h);
}
}node[maxn];
struct Tree
{
int sum,xx[25];
}tree[maxn<<2];
int n,m,cnt,top[maxn],ans[maxn],x,cc,FB[maxn];
bool cmp(Node a,Node b)
{
if(a.x!=b.x) return a.x<b.x;
if(a.y!=b.y) return a.y<b.y;
return a.flag<b.flag;
}
void push_up(int dex)
{
int k=0;
for(int i=1;i<=tree[dex<<1].sum;i++)
tree[dex].xx[++k]=tree[dex<<1].xx[i];
for(int i=1;i<=tree[dex<<1|1].sum;i++)
tree[dex].xx[++k]=tree[dex<<1|1].xx[i];
tree[dex].sum=k;
sort(tree[dex].xx+1,tree[dex].xx+1+tree[dex].sum);
if(k>10) tree[dex].sum=10;
}
void buildtree(int dex,int l,int r)
{
tree[dex].sum=0;
memset(tree[dex].xx,0,sizeof(tree[dex].xx));
if(l==r) return;
int mid=(l+r)>>1;
buildtree(dex<<1,l,mid);
buildtree(dex<<1|1,mid+1,r);
}
void Update(int dex,int l,int r,int pos,int val)
{
if(l==r)
{
tree[dex].xx[++tree[dex].sum]=val;
sort(tree[dex].xx+1,tree[dex].xx+1+tree[dex].sum);
if(tree[dex].sum>10) tree[dex].sum=10;
return;
}
int mid=(l+r)>>1;
if(pos<=mid) Update(dex<<1,l,mid,pos,val);
else Update(dex<<1|1,mid+1,r,pos,val);
push_up(dex);
}
void Query(int dex,int l,int r,int L,int R)
{
if(L<=l&&R>=r)
{
for(int i=1;i<=tree[dex].sum;i++)
FB[++cc]=tree[dex].xx[i];
sort(FB+1,FB+1+cc);
if(cc>10) cc=10;
return;
}
int mid=(l+r)>>1;
if(L<=mid) Query(dex<<1,l,mid,L,R);
if(R>mid) Query(dex<<1|1,mid+1,r,L,R);
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
cnt=0;
for(int i=1;i<=n+m;i++)
{
node[i].init();
node[i].id=i;
top[++cnt]=node[i].y;
if(i<=n) node[i].flag=0;
else node[i].flag=1;
}
sort(node+1,node+1+n+m,cmp);
sort(top+1,top+1+cnt);
x=unique(top+1,top+1+cnt)-(top+1);
buildtree(1,1,x);
for(int i=1;i<=n+m;i++)
{
int pos=lower_bound(top+1,top+1+x,node[i].y)-top;
if(!node[i].flag)
Update(1,1,x,pos,node[i].h);
else
{
cc=0;
Query(1,1,x,1,pos);
if(cc<node[i].h) ans[node[i].id]=-1;
else ans[node[i].id]=FB[node[i].h];
}
}
for(int i=n+1;i<=n+m;i++)
printf("%d\n",ans[i]);
}
return 0;
}