poj2010 Moo University - Financial Aid 优先队列
Description
Not wishing to admit dumber-than-average cows, the founders created an incredibly precise admission exam called the Cow Scholastic Aptitude Test (CSAT) that yields scores in the range 1..2,000,000,000.
Moo U is very expensive to attend; not all calves can afford it.In fact, most calves need some sort of financial aid (0 <= aid <=100,000). The government does not provide scholarships to calves,so all the money must come from the university's limited fund (whose total money is F, 0 <= F <= 2,000,000,000).
Worse still, Moo U only has classrooms for an odd number N (1 <= N <= 19,999) of the C (N <= C <= 100,000) calves who have applied.Bessie wants to admit exactly N calves in order to maximize educational opportunity. She still wants the median CSAT score of the admitted calves to be as high as possible.
Recall that the median of a set of integers whose size is odd is the middle value when they are sorted. For example, the median of the set {3, 8, 9, 7, 5} is 7, as there are exactly two values above 7 and exactly two values below it.
Given the score and required financial aid for each calf that applies, the total number of calves to accept, and the total amount of money Bessie has for financial aid, determine the maximum median score Bessie can obtain by carefully admitting an optimal set of calves.
Input
* Lines 2..C+1: Two space-separated integers per line. The first is the calf's CSAT score; the second integer is the required amount of financial aid the calf needs
Output
Sample Input
3 5 70
30 25
50 21
20 20
5 18
35 30
Sample Output
35
Hint
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
struct node{
int b,d;
}a[100005];
int cmp(struct node x,struct node y){
if(x.b==y.b) return x.d<y.d;
else return x.b>y.b;
}
int main(){
int n,c,f;
cin>>n>>c>>f;
for(int i=0;i<c;i++){
scanf("%d%d",&a[i].b,&a[i].d);
}
sort(a,a+c,cmp);
int plug=0;
//for(int i=0;i<c;i++)
///printf("%d %d\n",a[i].b,a[i].d);
if(n==1){
for(int i=0;i<c;i++){
if(a[i].d<=f){
cout<<a[i].b<<endl;
plug=1;
break;
}
}
}
else
for(int i=n/2;i<=c-1-n/2;i++){
priority_queue<int ,vector<int>,greater<int> >q;//前面
priority_queue<int ,vector<int>,greater<int> >p;//后面
for(int j=0;j<i;j++)
q.push(a[j].d);
int cost=0;
for(int j=0;j<n/2;j++)
cost+=q.top(),q.pop();
for(int j=i+1;j<c;j++)
p.push(a[j].d);
for(int j=0;j<n/2;j++)
cost+=p.top(),p.pop();
if(cost+a[i].d<=f){
cout<<a[i].b<<endl;
plug=1;
//cout<<1<<" "<<cost<<endl;
break;
}
//cout<<1<<" "<<cost<<endl;
if(plug==0) cout<<"-1"<<endl;
}
return 0;
}
这么写,会有很多次不必要的push()操作,使得超时,其中push()操作是n^2级别的;
真正的AC代码,push()操作是n级别的:
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
struct node{
int score,money;
}cow[100005];
int before[100005],after[100005];//分别表示前面部分的最小,和后面部分的最小,包括i本身
int cmp(struct node a,struct node b){
if(a.score==b.score) return a.money<b.money;
else return a.score>b.score;
}
int main(){
int n,c,f;
cin>>n>>c>>f;
for(int i=0;i<c;i++)
scanf("%d%d",&cow[i].score,&cow[i].money);
sort(cow,cow+c,cmp);
int sum=0;
priority_queue<int >q;
for(int i=0;i<n/2;i++) q.push(cow[i].money),sum+=cow[i].money;
before[n/2-1]=sum;
for(int i=n/2;i<=c-1-n/2;i++){
sum+=cow[i].money;
q.push(cow[i].money);
sum-=q.top();
q.pop();
before[i]=sum;
}
sum=0;
while(!q.empty()) q.pop();
for(int i=c-1;i>c-1-n/2;i--) q.push(cow[i].money),sum+=cow[i].money;
after[c-n/2]=sum;
for(int i=c-1-n/2;i>n/2;i--) {
sum+=cow[i].money;
q.push(cow[i].money);
sum-=q.top();
q.pop();
after[i]=sum;
}
int plug=0;
for(int i=n/2;i<=c-1-n/2;i++){
if(before[i-1]+after[i+1]+cow[i].money<=f){
plug=1;
cout<<cow[i].score<<endl;
break;
}
}
if(!plug) cout<<"-1"<<endl;
return 0;
}
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