Salary Changing CF-1251D（二分）

题意：

有$n$个员工，$s$元钱，现在要给每个员工发工资。每个员工的工资的范围$(l_i,r_i)$，求所有员工的工资中位数的最大值。

思路：

二分答案，$check$的时候判断工资可以大于等于$mid$的员工个数，用最小代价购买之后判断总价钱会不会超出范围。

代码：

 1 //#include<bits/stdc++.h>
 2 #include <set>
 3 #include <map>
 4 #include <stack>
 5 #include <cmath>
 6 #include <queue>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstring>
11 #include <iostream>
12 #include <algorithm>
13 
14 #define ll long long
15 #define pll pair<ll,ll>
16 #define pii pair<int,int>
17 #define bug printf("*********\n")
18 #define FIN freopen("input.txt","r",stdin);
19 #define FON freopen("output.txt","w+",stdout);
20 #define IO ios::sync_with_stdio(false),cin.tie(0)
21 #define ls root<<1
22 #define rs root<<1|1
23 #define pb push_back
24 
25 using namespace std;
26 const int inf = 2e9 + 7;
27 const ll Inf = 1e18 + 7;
28 const int maxn = 2e5 + 5;
29 const int mod = 1e9 + 7;
30 
31 struct node
32 {
33     ll l, r;
34 }p[maxn];
35 
36 ll s, n;
37 
38 bool cmp(const node& a, const node& b)
39 {
40     return a.l < b.l;
41 }
42 
43 bool check(ll x)
44 {
45     ll res = 0, m = (n + 1) >> 1;
46     for (int i = n; i >= 1; --i)
47     {
48         if (m && p[i].l <= x && p[i].r >= x)
49         {
50             res += x;
51             m--;
52         }
53         else
54         {
55             res += p[i].l;
56             if (p[i].l >= x && m)    m--;
57         }
58     }
59     if (m || res > s)    return 0;
60     return 1;
61 }
62 
63 
64 int main()
65 {
66     int T;
67     scanf("%d", &T);
68     while (T--)
69     {
70         scanf("%lld %lld", &n, &s);
71         for (int i = 1; i <= n; ++i)
72         {
73             scanf("%lld %lld", &p[i].l, &p[i].r);
74         }
75         sort(p + 1, p + 1 + n, cmp);
76         ll l = 0, r = s;
77         while (l <= r)
78         {
79             ll mid = (l + r) >> 1;
80             if (check(mid))    l = mid + 1;
81             else r = mid - 1;
82         }
83         printf("%lld\n", l - 1);
84     }
85 }