山东省第一届ACM大学生程序设计竞赛(原题) Ivan comes again!

Ivan comes again!

Time Limit: 1000MS Memory limit: 65536K

题目描述

The Fairy Ivan gave Saya three problems to solve (Problem F). After Saya finished the first problem (Problem H), here comes the second.
This is the enhanced version of Problem H.
There is a large matrix whose row and column are less than or equal to 1000000000. And there are three operations for the matrix:

1）add: Mark an element in the matrix. The element wasn’t marked before it is marked.
2）remove: Delete an element’s mark. The element
was marked before the element’s mark is deleted.
3）find: Show an element’s row and column,
and return a marked element’s row and column, where the marked element’s row and column are larger than the showed element’s row and column respectively. If there are multiple solutions, return the element whose row is the smallest; and if there are still
multiple solutions, return the element whose column is the smallest. If there is no solution, return -1.
Of course, Saya comes to you for help again.

输入

The input consists of several test cases.
The first line of input in each test case contains one integer
N (0<N≤200000), which represents the number of operations.
Each of the next
N lines containing an operation, as described above.
The last case is followed by a line containing one zero.

输出

For each case, print the case number (1, 2 …) first. Then, for each “find” operation, output the result.
Your output format should imitate the sample output. Print a blank line after each test case.

示例输入

4
add 2 3
find 1 2
remove 2 3
find 1 2

0

示例输出

Case 1:
2 3
-1

提示

 

来源

 2010年山东省第一届ACM大学生程序设计竞赛

示例程序

有3种操作

①往集合种加入一个点

②删除集合中一个点

③查询集合中是否存在大于该点的点有输出没有输出-1

根据set自动排序我们采用set来存点

点有二个元素组成所以使用pair

ACcode：

#include <set>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define maxn 100
using namespace std;

int main(){
int n,loop=1;
char str[10];
while(~scanf("%d",&n)&&n){
pair<int ,int >my;
set<pair<int ,int> >mm;
printf("Case %d:\n",loop++);
while(n--){
scanf("%s",&str);
scanf("%d%d",&my.first,&my.second);
if(str[0]=='a'){
mm.insert(my);
continue;
}
else if(str[0]=='r'){
mm.erase(my);
continue;
}
else {
set<pair <int,int> >::iterator id=mm.lower_bound(my);
while(id!=mm.end()){
if(id->first>my.first&&id->second>my.second){
printf("%d %d\n",id->first,id->second);
break;
}
id++;
}
if(id==mm.end())printf("-1\n");
}
}
printf("\n");
}
return 0;
}
/*
4
add 2 3
find 1 2
remove 2 3
find 1 2
*/