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POJ 2010 Moo University - Financial Aid
题目大意,从C头申请读书的牛中选出N头,这N头牛的需要的额外学费之和不能超过F,并且要使得这N头牛的中位数最大.若不存在,则输出-1(一开始因为没看见这个,wa了几次).
这个题的第一种做法就是用两个优先队列+贪心.
/* * Created: 2016年03月27日 14时41分47秒 星期日 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <algorithm> #include <string> #include <vector> #include <deque> #include <list> #include <set> #include <map> #include <stack> #include <queue> #include <numeric> #include <iomanip> #include <bitset> #include <sstream> #include <fstream> using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++) #define per(i,a,n) for (int i=n-1;i>=a;i--) #define inll(n) scanf("%I64d",&(n)) #define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2)) #define inlld(n) scanf("%lld",&(n)) #define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2)) #define inf(n) scanf("%f",&(n)) #define inf2(x1,x2) scanf("%f%f",&(x1),&(x2)) #define inlf(n) scanf("%lf",&(n)) #define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2)) #define inc(str) scanf("%c",&(str)) #define ins(str) scanf("%s",(str)) #define out(x) printf("%d\n",(x)) #define out2(x1,x2) printf("%d %d\n",(x1),(x2)) #define outf(x) printf("%f\n",(x)) #define outlf(x) printf("%lf\n",(x)) #define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2)); #define outlld(x) printf("%lld\n",(x)) #define outc(str) printf("%c\n",(str)) #define pb push_back #define mp make_pair #define fi first #define se second #define SZ(x) ((int)(x).size()) #define mem(X,Y) memset(X,Y,sizeof(X)); typedef vector<int> vec; typedef long long ll; typedef pair<int,int> P; const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1}; const int INF=0x3f3f3f3f; const ll mod=1e9+7; ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} const bool AC=true; struct point{ ll a; ll b; }; point p[100010]; ll l[100005]; ll r[100005]; bool cmp(point x,point y){ return x.a>y.a; } int main() { ll c,n,ans,sum,sum1,sum2,f; while(scanf("%lld %lld %lld",&c,&n,&f)==3){ rep(i,0,n) inlld2(p[i].a,p[i].b); sort(p,p+n,cmp); sum1=sum2=0;sum=0; priority_queue <int> que1; priority_queue <int> que2; mem(l,0); mem(r,0); rep(i,0,n){ //遍历数组,找出left[i],right[i]; if(i<c/2){ //i左边的数少于c/2时,全部加入队列 sum1+=p[i].b; que1.push(p[i].b); } else{ l[i]=sum1; if(p[i].b<que1.top()){ sum1-=que1.top(); //更新最小值 sum1+=p[i].b; que1.pop(); que1.push(p[i].b); } } } per(i,0,n){ if(i>=n-c/2){ sum2+=p[i].b; que2.push(p[i].b); } else{ r[i]=sum2; if(p[i].b<que2.top()){ sum2-=que2.top(); sum2+=p[i].b; que2.pop(); que2.push(p[i].b); } } } ans=-1; rep(i,c/2,n-c/2){ sum=p[i].b+l[i]+r[i]; if(sum<=f){ ans=p[i].a; break; } } outlld(ans); } return 0; }
这一题的第二种做法是二分
POJ 2010 Moo University - Financial Aid( 优先队列+二分查找)
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原文地址:http://www.cnblogs.com/akrusher/p/5349665.html