按第一关键字排序后枚举中位数，就变成了判断“左边前K小的和 + 这个中位数 + 右边前K小的和 <= F"，其中维护前K小和可以用treap做到。

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;struct node{    node *ch[2];    int sz;    int v;    int r;    int sum;    node(int v = 0) : v(v)    {        r = rand();        sz = 1;        ch[0] = ch[1] = NULL;        sum = v;    }    int cmp(int k)    {        return k > v;    }    void maintain()    {        sz = 1;        sum = v;        if(ch[0] != NULL)        {            sz += ch[0]->sz;            sum += ch[0]->sum;        }        if(ch[1] != NULL)        {            sz += ch[1]->sz;            sum += ch[1]->sum;        }    }};void rotate(node *&o, int d){    node *k = o->ch[d^1];    o->ch[d^1] = k->ch[d];    k->ch[d] = o;    o->maintain();    k->maintain();    o = k;}void insert(node *&o, int v){    if(o == NULL) o = new node(v);    else    {        int d = o->cmp(v);        insert(o->ch[d], v);        if(o->ch[d]->r > o->r) rotate(o, d^1);    }    o->maintain();}void remove(node *&o, int v){    if(v == o->v)    {        if(o->ch[0] == NULL) o = o->ch[1];        else if(o->ch[1] == NULL) o = o->ch[0];        else        {            int d = o->ch[0]->r < o->ch[1]->r ? 0 : 1;            rotate(o, d);            remove(o->ch[d], v);        }    } else    {        int d = o->cmp(v);        remove(o->ch[d], v);    }    if(o != NULL) o->maintain();}int query(node *o, int k){    int s = o->ch[0] == NULL ? 0 : o->ch[0]->sz;    int sum = o->ch[0] == NULL ? 0 : o->ch[0]->sum;    if(s + 1 > k) return query(o->ch[0], k);    if(s + 1 == k) return sum + o->v;    if(s + 1 < k) return sum + o->v + query(o->ch[1], k - s - 1);}void del(node *o){    if(o->ch[0] != NULL) del(o->ch[0]);    if(o->ch[1] != NULL) del(o->ch[1]);    delete o;}int main(){    int n, c, f;    while(scanf("%d%d%d", &n, &c, &f) != EOF)    {        vector
           
            
              > a; for(int i = 0; i < c; i++) { int x, y; scanf("%d%d", &x, &y); a.push_back(make_pair(x, y)); } sort(a.begin(), a.end()); node *r = NULL, *l = NULL; for(int i = 0; i < n / 2; i++) insert(r, a[c - i - 1].second); for(int i = c - n / 2 - 1; i >= 0; i--) insert(l, a[i].second); int ans = -1; for(int i = c - n / 2 - 1; i >= n / 2; i--) { remove(l, a[i].second); if(query(l, n / 2) + query(r, n / 2) + a[i].second <= f) { ans = a[i].first; break; } insert(r, a[i].second); } printf("%d\n", ans); del(l); del(r); } return 0;}