cond clu ref console padding top ostream name consola

題目鏈接：

huangjing

題意：給了三種操作
1：add（x,y）將這個點增加二維坐標系
2：remove（x,y）將這個點從二維坐標系移除。
3：find（x，y）就是找到在（x，y）右上方的第一個點。
思路：我們能夠建立n個set以x為橫坐標，那麽我們這個題就轉化為找一個最小的x是否存在滿足條件，那麽x一旦被找到。那麽縱坐標就自然而然的找到了。當然更新操作就是對maxy的維護，然後查詢操作就是找出一個最小的x。。

還有由於n很大，所以要採用離散化的方法。然後進行離線處理。還是就是掌握set的使用方法，看來得買本c++primer了。insert（int val）將val插入到set中。且保證有序，erase（int val）將val刪除且保證有序。。

題目：

D. Points time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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 test(s) input

7
add 1 1
add 3 4
find 0 0
remove 1 1
find 0 0
add 1 1
find 0 0

output

1 1
3 4
1 1

input

13
add 5 5
add 5 6
add 5 7
add 6 5
add 6 6
add 6 7
add 7 5
add 7 6
add 7 7
find 6 6
remove 7 7
find 6 6
find 4 4

output

7 7
-1
5 5

代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=200000+10;
struct Tree
{
    int x,y;
    char op[5];
    void init()
    {
        scanf("%s%d%d",op,&x,&y);
    }
}tree[maxn<<2];

set<int>xx[maxn];
int maxy[maxn<<2],top[maxn],n,m;

void buildtree(int l,int r,int dex)
{
    maxy[dex]=-1;
    if(l==r) return;
    int mid=(l+r)>>1;
    buildtree(l,mid,dex<<1);
    buildtree(mid+1,r,dex<<1|1);
}

void update(int l,int r,int dex,int pos)
{
     if(l==r)
     {
         if(xx[pos].size())
            maxy[dex]=*(--xx[pos].end());
         else
            maxy[dex]=-1;
         return;
     }
     int mid=(l+r)>>1;
     if(pos<=mid)  update(l,mid,dex<<1,pos);
     else  update(mid+1,r,dex<<1|1,pos);
     maxy[dex]=max(maxy[dex<<1],maxy[dex<<1|1]);
}

int Query(int l,int r,int dex,int L,int R,int val)
{
   if(maxy[dex]<=val||L>R)  return -1;//這裏L>R是一個trick，wa在了第4組數據，有可能L>R
   if(l==r) return l;
   int mid=(l+r)>>1;
   if(R<=mid)  return Query(l,mid,dex<<1,L,R,val);
   else if(L>mid)  return Query(mid+1,r,dex<<1|1,L,R,val);
   else
   {
       int t=Query(l,mid,dex<<1,L,R,val);
       if(t!=-1)  return t;
       return Query(mid+1,r,dex<<1|1,L,R,val);
   }
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            tree[i].init();
            top[i]=tree[i].x;
        }//離散化
        sort(top+1,top+1+n);
        m=unique(top+1,top+1+n)-(top+1);
        for(int i=1;i<=m;i++)
            xx[i].clear();
        buildtree(1,m,1);
        for(int i=1;i<=n;i++)
        {
             int pos=lower_bound(top+1,top+1+m,tree[i].x)-top;
             if(tree[i].op[0]==‘a‘)
             {
                 xx[pos].insert(tree[i].y);
                 update(1,m,1,pos);
             }
             else if(tree[i].op[0]==‘r‘)
             {
                 xx[pos].erase(tree[i].y);
                 update(1,m,1,pos);
             }
             else
             {
                int ans=Query(1,m,1,pos+1,m,tree[i].y);
                if(ans==-1)
                    printf("-1\n");
                else
                    printf("%d %d\n",top[ans],*xx[ans].upper_bound(tree[i].y));
             }
        }
    }
    return 0;
}

CodeForces 19D Points（離散化+線段樹+單點更新）