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codeforces 19D D. Points 树套树

D. Points

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/19/problem/D

Description

Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0, 0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:

  • add x y — on the sheet of paper Bob marks a point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is not yet marked on Bob's sheet at the time of the request.
  • remove x y — on the sheet of paper Bob erases the previously marked point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is already marked on Bob's sheet at the time of the request.
  • find x y — on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x, y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.

Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2·105, Bob failed to cope. Now he needs a program that will answer all Pete's requests. Help Bob, please!

Input

The first input line contains number n (1 ≤ n ≤ 2·105) — amount of requests. Then there follow n lines — descriptions of the requests. add x y describes the request to add a point, remove x y — the request to erase a point, find x y — the request to find the bottom-left point. All the coordinates in the input file are non-negative and don't exceed 109.

Output

For each request of type find x y output in a separate line the answer to it — coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x, y). If there are no points strictly above and to the right of point (x, y), output -1.
 

Sample Input

7

add 1 1

add 3 4

find 0 0

remove 1 1

find 0 0

add 1 1

find 0 0


Sample Output

1 1

3 4

1 1

HINT

 

题意

在一个直角坐标系内，每次有三种操作

１．插入一个点(x,y)保证(x,y)未插入过

２．删除一个点(x,y)保证(x,y)插入过

３．询问点(x,y)右上角所有点最靠左边，然后最靠下面的点的位置

题解：

我们可以用set降维，就是每个叶子几点有一个set存y值，然后我们可以记录一个max，存最大的y值，这样对于每次询问(x,y),问题转化成求x点右边第一个max比y大的位置，

然后再用lower_bound求set中刚好大于它的y值。感觉好像也不是很难。

代码：

 

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define N 500050
  4 int n,cnt,kth[N];
  5 struct Query{int l,r;char id;}que[N];
  6 struct Tree{int l,r,max;set<int>sx;}tr[N<<2];
  7 template<typename T>void read(T&x)
  8 {
  9   int k=0;char c=getchar();
 10   x=0;
 11   while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
 12   if (c==EOF)exit(0);
 13   while(isdigit(c))x=x*10+c-'0',c=getchar();
 14   x=k?-x:x;
 15 }
 16 
 17 void read_char(char &c)
 18 {while(!isalpha(c=getchar()));};
 19 void push_up(int x)
 20 {
 21   tr[x].max=max(tr[x<<1].max,tr[x<<1|1].max);
 22 }
 23 void bt(int x,int l,int r)
 24 {
 25   //tr[x]=Tree{l,r,0};
 26   tr[x].max=0;
 27   tr[x].l=l; tr[x].r=r;
 28   if (l==r)return;
 29   int mid=(l+r)>>1;
 30   bt(x<<1,l,mid);
 31   bt(x<<1|1,mid+1,r);
 32 }
 33 void add(int x,int p,int tt)
 34 {
 35   if (tr[x].l==tr[x].r)
 36     {
 37       tr[x].sx.insert(tt);
 38       tr[x].max=max(tr[x].max,tt);
 39       return;
 40     }
 41   int mid=(tr[x].l+tr[x].r)>>1;
 42   if (p<=mid)add(x<<1,p,tt);
 43   if (mid<p)add(x<<1|1,p,tt);
 44   push_up(x);
 45 }
 46 void del(int x,int p,int tt)
 47 {
 48   if (tr[x].l==tr[x].r)
 49     {
 50       tr[x].sx.erase(tt);
 51       tr[x].max=tr[x].sx.empty()?0:*(--tr[x].sx.end());
 52       return;
 53     }
 54   int mid=(tr[x].l+tr[x].r)>>1;
 55   if (p<=mid)del(x<<1,p,tt);
 56   if (mid<p)del(x<<1|1,p,tt);
 57   push_up(x);
 58 }
 59 pair<int,int> query(int x,int p,int tt)
 60 {
 61   if (tr[x].max<tt)return make_pair(-1,-1);
 62   if (tr[x].l==tr[x].r)
 63     return make_pair(tr[x].l,*(tr[x].sx.lower_bound(tt)));
 64   int mid=(tr[x].l+tr[x].r)>>1;
 65   if (p>mid)return query(x<<1|1,p,tt);
 66   else
 67     {
 68       pair<int,int> tp=query(x<<1,p,tt);
 69       return tp.first==-1?query(x<<1|1,p,tt):tp;
 70     }
 71 }
 72 int main()
 73 {
 74   #ifndef ONLINE_JUDGE
 75   freopen("aa.in","r",stdin);
 76   #endif
 77   read(n);
 78   for(int i=1;i<=n;i++)
 79     {
 80       read_char(que[i].id);
 81       read(que[i].l);
 82       read(que[i].r);
 83       kth[++cnt]=que[i].l;
 84       kth[++cnt]=que[i].r;
 85     }
 86   sort(kth+1,kth+cnt+1);
 87   cnt=unique(kth+1,kth+cnt+1)-kth-1;
 88   bt(1,1,cnt);
 89   for(int i=1;i<=n;i++)
 90     {
 91       int p=lower_bound(kth+1,kth+cnt+1,que[i].l)-kth;
 92       int tt=lower_bound(kth+1,kth+cnt+1,que[i].r)-kth;
 93       if (que[i].id=='a')add(1,p,tt);
 94       if (que[i].id=='r')del(1,p,tt);
 95       if (que[i].id=='f')
 96     {
 97       pair<int,int> ans=query(1,p+1,tt+1);
 98       if (ans.first==-1)printf("-1\n");
 99       else 
100       printf("%d %d\n",kth[ans.first],kth[ans.second]);
101     }
102     }
103 }
View Code

 

posted @ 2019-04-22 23:35  mmqqdd  阅读(210)  评论(0)  编辑  收藏  举报