POJ 2010 (Priority queue

Sweep with priority queue to get the minimum value of the sum of the values greater than and less than the median, and sweep again to get the optimal solution

#include <iostream>#include<cstdio>#include<algorithm>#include<queue>#include<utility>#include<vector>#defineINF 0X3FFFFFFFusing namespaceStd;typedefLong Longll;intN,c,f;typedef pair<int,int>PII;Const intmaxv=1e5+ -;p Air<int,int> COW[MAXV];////score,aidll LOW[MAXV],UPP[MAXV];BOOLCMP (PII a,pii b) {returna.second<B.second;}intMain () {Freopen ("In.txt","R", stdin); CIN>>N>>C>>F;  for(intI=0; i<c;i++) scanf ("%d%d",&cow[i].first,&Cow[i].second); Sort (Cow,cow+b); ll Tol=0; Priority_queue<int>q1,q2; ll half=n/2;  for(intI=0; i<c;i++) {Low[i]= (Q1.size () ==half?Tol:inf);        Q1.push (Cow[i].second); Tol+=Cow[i].second; if(Q1.size () >half) {Tol-=Q1.top ();        Q1.pop (); }} Tol=0;  for(inti=c-1; i>=0; i--) {Upp[i]= (Q2.size () ==half?Tol:inf);        Q2.push (Cow[i].second); Tol+=Cow[i].second; if(Q2.size () >half) {Tol-=Q2.top ();        Q2.pop (); }    }    intans=-1;  for(intI=0; i<c;i++){        if(low[i]+upp[i]+cow[i].second<=F) ans=Cow[i].first; } cout<<ans<<Endl; return 0;}

POJ 2010 (Priority queue