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 longconst 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;}