链接：
https://codeforces.com/contest/1251/problem/D
题意：
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 l1+l2+⋯+ln≤s.
Note that you don't have to spend all your s dollars on salaries.
You have to answer t test cases.
思路：
二分， 判断赛的时候贪心判断。
代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e5+10;
struct Node
{
int l, r;
bool operator < (const Node& rhs) const
{
if (this->l != rhs.l)
return this->l > rhs.l;
return this->r > rhs.r;
}
}node[MAXN];
int n;
LL s;
bool Check(LL val)
{
int p = n/2+1;
LL sum = 0;
for (int i = 1;i <= n;i++)
{
if (node[i].r >= val && p > 0)
{
sum += max(val, (LL)node[i].l);
p--;
}
else
{
sum += node[i].l;
}
}
return (sum <= s && p == 0);
}
int main()
{
ios::sync_with_stdio(false);
int t;
cin >> t;
while(t--)
{
cin >> n >> s;
for (int i = 1;i <= n;i++)
cin >> node[i].l >> node[i].r;
sort(node+1, node+1+n);
LL l = 1, r = s;
LL res = 0;
while(l <= r)
{
LL mid = (l+r)/2;
//cout << mid << endl;
if (Check(mid))
{
res = max(res, mid);
l = mid+1;
}
else
r = mid-1;
}
cout << res << endl;
}
return 0;
}