最大化最小甜度问题适合用二分法求解。通过二分猜测最小甜度 $mid$，然后验证是否存在至少 $k$ 种水果的甜度 $≥ mid$，且其中价格最小的 $k$ 种水果的总价格 $≤ B$。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
const int N = 1e5 + 10; 
struct Fruit { 
    int sweetness; 
    int price;     
} fruits[N]; 

int n, B, k; 
bool cmp(Fruit a, Fruit b) { 
    return a.price < b.price; 
}

bool check(int mid) { 
    vector<Fruit> valid;
    for (int i = 1; i <= n; i++) { 
        if (fruits[i].sweetness >= mid) {
            valid.push_back(fruits[i]);
        }
    }
    int cnt = valid.size();
    if (cnt < k) return false;  
    sort(valid.begin(), valid.end(), cmp);
    int sum = 0;
    for (int i = 0; i < k; i++) { 
        sum += valid[i].price;
        if (sum > B) return false; 
    }
    return sum <= B; 
}

void solve() { 
    cin >> n >> B >> k;
    int min_sweet = INT_MAX, max_sweet = 0;
    for (int i = 1; i <= n; i++) { 
        cin >> fruits[i].sweetness >> fruits[i].price;
        min_sweet = min(min_sweet, fruits[i].sweetness); 
        max_sweet = max(max_sweet, fruits[i].sweetness); 
    }
    
    int left = min_sweet, right = max_sweet;
    int ans = -1; 
    
    while (left <= right) { 
        int mid = left + (right - left) / 2; 
        if (check(mid)) { 
            ans = mid; 
            left = mid + 1; 
        } else { 
            right = mid - 1; 
        }
    }
    
    cout << ans << endl; 
}

signed main() { 
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    solve();
    return 0;
}