CodeForces-1251D Salary Changing

传送门：CodeForces-1251D

思路分析

二分中位数是哪个
将所有人按l排序，从最后开始枚举，贪心的分发工资

样例输入

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

样例输出

11
1337
6

AC代码

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define int long long
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
using namespace __gnu_pbds;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree;

const int INF = 0x3f3f3f3f;
const int N=2e5+10;

pair<int,int>a[N];

signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int t;
    read(t);
    while(t--){
        int n,s;
        read(n,s);
        for(int i=1;i<=n;i++) read(a[i].first,a[i].second);
        sort(a+1,a+1+n);
        int l=0,r=s+1;
        int ans;
        while(r>=l){
            int mid=l+r>>1;
            int cnt=0;
            int sum=0;
            for(int i=n;i>=1;i--){
                if(a[i].second>=mid && cnt<(n+1)/2){
                    cnt++;
                    sum+=max(mid,a[i].first);
                }else{
                    sum+=a[i].first;
                }
            }
            if(cnt==(n+1)/2 && sum<=s) {
                ans=mid;
                l=mid+1;
            }else{
                r=mid-1;
            }
        }
        printf("%lld\n",ans);
    } 




    return 0;
}