链接
https://codeforces.com/contest/1251/problem/D
题目
You are the head of a large enterprise. n people work at you, and n is odd (i. e. n is not divisible by 2).
You have to distribute salaries to your employees. Initially, you have s dollars for it, and the i-th employee should get a salary from li to ri dollars. You have to distribute salaries in such a way that the median salary is maximum possible.
To find the median of a sequence of odd length, you have to sort it and take the element in the middle position after sorting. For example:
the median of the sequence [5,1,10,17,6] is 6,
the median of the sequence [1,2,1] is 1.
It is guaranteed that you have enough money to pay the minimum salary, i.e (l_1+l_2+⋯+l_n≤s.)
Note that you don't have to spend all your s dollars on salaries.
You have to answer t test cases.
Input
The first line contains one integer t ((1≤t≤2⋅10^5)) — the number of test cases.
The first line of each query contains two integers n and s ((1≤n<2⋅10^5,1≤s≤2⋅10^{14})) — the number of employees and the amount of money you have. The value n is not divisible by 2.
The following n lines of each query contain the information about employees. The i-th line contains two integers li and ri ((1≤li≤ri≤10^9)).
It is guaranteed that the sum of all n over all queries does not exceed (2⋅10^5).
It is also guaranteed that you have enough money to pay the minimum salary to each employee, i.e.(sum_{i=1}^nl_i≤s)
Output
For each test case print one integer — the maximum median salary that you can obtain.
Example
input
3
3 26
10 12
1 4
10 11
1 1337
1 1000000000
5 26
4 4
2 4
6 8
5 6
2 7
outputCopy
11
1337
6
Note
In the first test case, you can distribute salaries as follows: sal1=12,sal2=2,sal3=11 (sali is the salary of the i-th employee). Then the median salary is 11.
In the second test case, you have to pay 1337 dollars to the only employee.
In the third test case, you can distribute salaries as follows: (sal_1=4,sal_2=3,sal_3=6,sal_4=6,sal_5=7.) Then the median salary is 6.
思路
二分枚举中位数(k)，遍历一遍所有员工，对于(r_i)小于(k)的人发放(l_i)元,(cntl++)，对于(l_i)大于k的人发放(l_i)元,(cntr++),然后剩下的人当中取(n/2-cntl)，发放(l_i)，使得他们薪水是小于(k)的，最后剩下人发放(k)。以及判断大于和小于(k)的人数是否合法还有发放总钱数不能大于(S)就不说了。
然后这道题的需要注意二分的范围，并不是在负无穷和正无穷之间具有单调性，例如样例1，(k)取(1)就不合法，但答案可以是(6).可以发现答案在(a[n/2+1].l)到正无穷具有单调性（或者说在(a[n/2+1].l)(以(l)为第一关键字排序)到(a[n/2+1].r)(以(r)为第一关键字排序)具有单调性）。(k=a[n/2+1].l)时，可以有(n/2)个人小于等于(k),(n/2)个人大于(k)。
代码
#include<bits/stdc++.h>
#define x first
#define y second
#define int long long
using namespace std;
typedef long long LL;
const int N=200010;
typedef pair<LL,LL> PII;
PII a[N];
int n;
LL s,tmp[N];
vector<LL> v;
bool check(LL k){
int ls=0,rs=0,cnt=0;
LL sum=0;
for(int i=1;i<=n;++i){
if(a[i].y<k) ls++,sum+=a[i].x;
else if(a[i].x>k) rs++,sum+=a[i].x;
else tmp[++cnt]=a[i].x;
}
if(sum>s||ls>n/2||rs>n/2) return false;
for(int i=1;i<=cnt;++i) {
if(ls==n/2) break;
sum+=tmp[i];
ls++;
}
sum+=(n/2+1-rs)*k;
return sum<=s;
}
signed main(){
int T;
scanf("%lld",&T);
while(T--){
scanf("%lld%lld",&n,&s);
for(int i=1;i<=n;++i)
scanf("%lld%lld",&a[i].x,&a[i].y);
sort(a+1,a+1+n);
LL l=a[n/2+1].x,r=1e14,ans;
while(l<=r){
LL mid=(l+r)/2;
if(check(mid)){
l=mid+1;
ans=mid;
}
else r=mid-1;
}
cout<<ans<<endl;
}
return 0;
}