Welcome Party
Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 875 Accepted Submission(s): 194
Problem Description
The annual welcome party of the Department of Computer Science and Technology is coming soon! Many students have been applying to show up at the welcome party, and every one of them can choose to sing a song or play crosstalk. This troubles the chief director a lot: how to arrange the program list, such that every student can have a chance to show up on the stage, and the satisfactory value of audiences is maximized?
To cope with this problem, the director proposes a model. In this model, every student has two attributes: the singing ability and crosstalking ability. The satisfactory value of audiences to singings is the maximum singing ability among all students that choose to sing a song; similarly, the satisfactory value to crosstalks is the maximum crosstalking ability among all students that choose play crosstalk. The strange thing is, the overall satisfactory value to the whole party is negatively related to the absolute difference between the satisfactory values to singings and crosstalks. The problem is, what is the minimum possible absolute difference between the satisfactory values of the two types of programs?
Note that:
- every student should choose exactly one type of programs to play;
- at least one student should sing a song, and at least one student should play crosstalk.
To cope with this problem, the director proposes a model. In this model, every student has two attributes: the singing ability and crosstalking ability. The satisfactory value of audiences to singings is the maximum singing ability among all students that choose to sing a song; similarly, the satisfactory value to crosstalks is the maximum crosstalking ability among all students that choose play crosstalk. The strange thing is, the overall satisfactory value to the whole party is negatively related to the absolute difference between the satisfactory values to singings and crosstalks. The problem is, what is the minimum possible absolute difference between the satisfactory values of the two types of programs?
Note that:
- every student should choose exactly one type of programs to play;
- at least one student should sing a song, and at least one student should play crosstalk.
Input
The first line of input consists of a single integer T (1≤T≤70), the number of test cases.
Each test case starts with a line of a single integer n (2≤n≤100000), denoting the number of students applying to show up on the stage. Then follow n lines, each containing two integers x and y (0≤x,y≤1018), denoting the singing ability and crosstalking ability of a student.
It is guaranteed that the sum of n over all test cases never exceeds 1000000.
Each test case starts with a line of a single integer n (2≤n≤100000), denoting the number of students applying to show up on the stage. Then follow n lines, each containing two integers x and y (0≤x,y≤1018), denoting the singing ability and crosstalking ability of a student.
It is guaranteed that the sum of n over all test cases never exceeds 1000000.
Output
For each test case, output a single integer, denoting the minimum possible absolute difference between the satisfactory values of the two types of programs.
Sample Input
2
5
27 46
89 13
55 8
71 86
22 35
3
3 5
4 7
6 2
Sample Output
3
1
Source
题意:
有n个学生，每个学生有x值和y值，每个学生必须要么进入x集合要么y集合，x集合的代表值是这个集合中最大的x值，y集合同理，问如何分配学生才能使得x集合代表值与y集合的代表值差值最小
思路:
n最大1e5，所以复杂度最大只能为O(nlogn),可以考虑遍历x集合的同时二分y集合
用一个结构体存学生的x值和y值，并对对学生按x和y降序排序，将学生的y值加入到multiset中
因为现在是降序排序，所以如果选了一个学生代表x集合的代表，则x比他大的学生都必须要加入到y集合中，而比小于等于他的那些则可以随意
所以我们枚举每个学生当x集合的代表，并将其在y集合中的值删掉，用maxn维护必须要加入y集合的那些值的最大值，multiset维护可能可以加入y集合的值，并multiset中二分一个和他最接近的值（可能大于，等于或小于，所以如果it!=s.end()还有it--判断)，
当然这个值要比()才能更新ans，再用maxn和现在这个学生x值的差值更新ans,这个学生判断完后就去判断下一个学生，当然因为我们对他们x值降序排序了，所以这个学生就必须加入y集合，用其y值更新maxn
注意(坑点,WA了几十次TAT):
1.maxn代表必定要分到另一个集合的元素的最大值，而第一次没有必须要分到另一个集合中的元素，所以第一次是不能调用的
2.因为一个集合的代表值是这个集合中所以元素中的最大值，所以二分到的值必须要大于maxn才能更新ans
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int amn=1e5+5; 5 struct node{ 6 ll x,y; 7 }a[amn]; 8 multiset<ll> s; 9 multiset<ll>::iterator it; 10 bool cmp(node a,node b){ 11 if(a.x==b.x)return a.y>b.y; 12 return a.x>b.x; 13 } 14 int main(){ 15 int n,T; 16 bool f; 17 ll ans,maxn; 18 scanf("%d",&T); 19 while(T--){ 20 scanf("%d",&n); 21 for(int i=1;i<=n;i++){ 22 scanf("%lld%lld",&a[i].x,&a[i].y); 23 s.insert(a[i].y); ///将y值插入multiset 24 } 25 sort(a+1,a+1+n,cmp); ///对学生按x降序排序 26 ans=1e18; 27 maxn=0; ///maxn维护一个必须要加入y集合的学生中y值最大的值 28 for(int i=1;i<=n;i++){ 29 s.erase(s.find(a[i].y)); ///因为现在把当前学生分到x集合中，所以删除当前学生在multiset中的y值 30 if(i>1)ans=min(ans,abs(a[i].x-maxn)); ///这里是关键，maxn代表必定要分到另一个集合的元素的最大值，而第一次没有必须要分到另一个集合中的元素，所以第一次是不能调用的，否则因为一开始maxn为0，所以第一次会将ans更新为a[1].x 31 it=s.lower_bound(a[i].x); ///在multiset中二分一个和当前学生x值最接近的值 32 if(it!=s.begin()&&it==s.end())it--; ///lower_bound会返回第一个大于等于搜索值的地址，如果不存在则返回s.end(),所以如果迭代器不是s.begin()且是s.end()则要将其指向上一个元素 33 if(*it>=maxn)ans=min(ans,abs(a[i].x-*it)); ///因为一个集合的代表值是这个集合中所以元素中的最大值，所以二分到的值必须要大于maxn才能更新ans 34 if(it!=s.begin()){ ///如果前面还有元素，则看上一个元素是否符合条件 35 it--; 36 if(*it>=maxn)ans=min(ans,abs(a[i].x-*it)); 37 } 38 maxn=max(maxn,a[i].y); ///现在这个学生必须要被加到y集合中了，更新一下maxn 39 } 40 printf("%lld ",ans); 41 } 42 } 43 /** 44 有n个学生，每个学生有x值和y值，每个学生必须要么进入x集合要么y集合，x集合的代表值是这个集合中最大的x值，y集合同理，问如何分配学生才能使得x集合代表值与y集合的代表值差值最小 45 n最大1e5，所以复杂度最大只能为O(nlogn),可以考虑遍历x集合的同时二分y集合 46 用一个结构体存学生的x值和y值，并对对学生按x和y降序排序，将学生的y值加入到multiset中 47 因为现在是降序排序，所以如果选了一个学生代表x集合的代表，则x比他大的学生都必须要加入到y集合中，而比小于等于他的那些则可以随意 48 所以我们枚举每个学生当x集合的代表，并将其在y集合中的值删掉，用maxn维护必须要加入y集合的那些值的最大值，multiset维护可能可以加入y集合的值，并multiset中二分一个和他最接近的值（可能大于，等于或小于，所以如果it!=s.end()还有it--判断)， 49 当然这个值要比()才能更新ans，再用maxn和现在这个学生x值的差值更新ans,这个学生判断完后就去判断下一个学生，当然因为我们对他们x值降序排序了，所以这个学生就必须加入y集合，用其y值更新maxn 50 注意(坑点): 51 1.maxn代表必定要分到另一个集合的元素，而第一次没有必须要分到另一个集合中的元素，所以第一次是不能调用的 52 2.因为一个集合的代表值是这个集合中所以元素中的最大值，所以二分到的值必须要大于maxn才能更新ans 53 **/