按照score排序,贪心,从左到右用堆维护并且记录前面的最小N/2个花费之和。
然后从右向左枚举中位数,维护N/2个数之和加上并判断是否满足条件。(stl的队列没有clear(),只能一个一个pop…
复杂度O(nlogn)
也可以二分。先按照score排序,记录备份牛在排序后的下标。然后将备份按照资金排序,
,二分中位数的值,对于mid值,按照备份顺序贪心选择牛,根据之前的下标判断在左边或者右边。
因为是最贪心的选,如果两边都不够,说明无解。如果某一边少了,mid就应该往相反的方向移动。
如果满足条件就记录答案,并向上缩小区间。
复杂度也是O(nlogn)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
//#include<bits/stdc++.h>
using namespace std;#define PB push_back
#define MP make_pair
#define fi first
#define se second
#define PS pushconst int maxn = 1e5+;
const int maxsz = 1e4+;
struct BinaryHEAP
{
int Heap[maxsz];
int sz;
#define Cmp(a,b) ((a)>(b))
void push(int x)
{
int i = ++sz;
while(i > ){
int p = i>>;
if(!Cmp(x,Heap[p])) break;
Heap[i] = Heap[p];
i = p;
}
Heap[i] = x;
} void pop()
{
int x = Heap[sz--];
int i = ;
while((i<<)<=sz){
int a = i<<, b = i<<|;
if(b<=sz && Cmp(Heap[b],Heap[a])) a = b;
if(!Cmp(Heap[a],x)) break;
Heap[i] = Heap[a];
i = a;
}
Heap[i] = x;
}
int top(){ return Heap[]; }
//int operator[](int x){ return Heap[x]; }
}q;pair<int,int> cow[maxn];
int N, C, F;
int Half;
int preSum[maxn];int sol()
{
sort(cow,cow+C);
int Half = N>>, sum = ;
for(int i = ; i < Half; i++) {
sum += cow[i].se;
q.push(cow[i].se);
}
for(int i = Half; i < C-Half; i++){
preSum[i] = sum;
if(q.top() > cow[i].se) {
sum -= q.top();
q.pop();
sum += cow[i].se;
q.push(cow[i].se);
} }
q.sz = ;
sum = ;
for(int i = C-Half; i < C; i++) {
sum += cow[i].se;
q.push(cow[i].se);
}
for(int i = C-Half; --i >= Half; ){
if(sum + preSum[i] + cow[i].se <= F) return cow[i].fi;
if(q.top() > cow[i].se) {
sum -= q.top();
q.pop();
sum += cow[i].se;
q.push(cow[i].se);
}
}
return -;
}//#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif scanf("%d%d%d",&N,&C,&F);
for(int i = ; i < C; i++){
scanf("%d%d",&cow[i].fi,&cow[i].se);
}
printf("%d\n",sol());
return ;
}