poj 2010

这道题目之前好像差不多做过一次，所以比较有思路。我们先将所有的值按照score从小到大排序，然后用优先队列(按照financial大的在前)维护某个数前面和后面各前n/2个最小的数，sum1[i]代表i前面n/2个最小的finacial之和，sum2[i]代表i后面n/2个最小的financial之和。最后从后往前遍历一遍，找到一个sum1[i]+sum2[i]+a[i].financial<=f的就输出，否则输出-1.

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
#include<set>
#include<bitset>
//#define ONLINE_JUDGE
#define eps 1e-5
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
#define ll long long
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
//#pragma comment(linker,"/STACK:1024000000,1024000000")
int n,m,L;
#define N 50005
#define M 1000100
#define Mod 1000000007
#define p(x,y) make_pair(x,y)
struct Node{
    int sc,fn;
    bool operator <(const Node &x)const{
        return sc<x.sc;
    }
}a[100010];
struct cmp1{
    bool operator ()(const Node &x,const Node &y){
        return x.fn<y.fn;
    }
};
ll sum1[100010];
ll sum2[100010];
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
#endif
    int c,f;
    while(scanf("%d%d%d",&n,&c,&f)!=EOF){
        for(int i=0;i<c;i++)
            sfd(a[i].sc,a[i].fn);
        sort(a,a+c);
        memset(sum1,0,sizeof sum1);
        memset(sum2,0,sizeof sum2);
        ll ans=0;
        int size = n/2;
        priority_queue<Node,vector<Node>,cmp1> pq1;
        for(int i=0;i<c;i++){
            sum1[i] = (int)pq1.size()==size?ans:0;
            ans += (ll)a[i].fn;
            pq1.push(a[i]);
            if((int)pq1.size()>size){
                ans -= pq1.top().fn;
                pq1.pop();
            }
        }
        priority_queue<Node,vector<Node>,cmp1> pq2;
        ans=0;
        for(int i=c-1;i>=0;i--){
            sum2[i] = (int)pq2.size()==size?ans:0;
            ans += (ll)a[i].fn;
            pq2.push(a[i]);
            if((int)pq2.size()>size){
                ans -= pq2.top().fn;
                pq2.pop();
            }
        }
        int flag=0;
        for(int i=c-1;i>=0;i--){
            if(sum1[i] && sum2[i]){
                if(sum1[i]+sum2[i]+a[i].fn<=f){
                    flag=1;
                    printf("%d\n",a[i].sc);
                    break;
                }
            }
        }
        if(!flag)
            printf("-1\n");
    }
    return 0;
}