poj2010 Moo University - Financial Aid Priority Queue

Description

Bessie noted that although humans have many universities they can attend, cows have none. To remedy this problem, she and her fellow cows formed a new university called The University of Wisconsin-Farmside,"Moo U" for short.

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

* Line 1: Three space-separated integers N, C, and F

* 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

* Line 1: A single integer, the maximum median score that Bessie can achieve. If there is insufficient money to admit N calves,output -1.

Sample Input

Sample Output

Hint

Sample output:If Bessie accepts the calves with CSAT scores of 5, 35, and 50, the median is 35. The total financial aid required is 18 + 30 + 21 = 69 <= 70.

Chinese meaning of problems: there are cows c, each cow has two attributes socre, money; n select from cows, n must be an odd number, so that the properties of the n money cows and F or less, this condition is satisfied under the premise that the n cows (median score attribute) as large as the median and outputs the median score, and if not satisfy prerequisites outputs 1;

Thinking: c cows will be sorted by score in descending properties, enumeration traversal to find the n / 2 cows (minimum attribute value before money n / 2 cows before i i-1 at the head of the cow ), similarly to identify n / 2 cows in the i-th cows, so that money and minimum. The first meet the first n / 2 cows money and money n / 2 money cows and cows plus the i after adding <= f, the output of the i-th cow can score; otherwise, the output 1;

Timeout Code:

#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;
}

So write, there will be many times unnecessarily push () operation, making the timeout, which push () operation is n ^ 2 levels;

The real AC codes, push () operation is n-level:

#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;
}

poj2010 Moo University - Financial Aid Priority Queue

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