題目鏈接:
huangjing
題意:給了三種操作
1:add(x,y)將這個點加入二維坐標系
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:我們可以建立n個set以x為橫坐標,那麼我們這個題就轉化為找一個最小的x是否存在滿足條件,那麼x一旦被找到,那麼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。。還有因為n非常大,所以要采用離散化的方法,然後進行離線處理。還是就是掌握set的用法,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 outputPete 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:
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!
InputThe 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.
OutputFor 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) input7 add 1 1 add 3 4 find 0 0 remove 1 1 find 0 0 add 1 1 find 0 0output
1 1 3 4 1 1input
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 4output
7 7 -1 5 5代碼:
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