Moo University - Financial Aid
| Time Limit: 1000MS |
|
Memory Limit: 30000K |
| Total Submissions: 3115 |
|
Accepted: 945 |
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
3 5 70
30 25
50 21
20 20
5 18
35 30
Sample Output
35
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.
Source
USACO 2004 March Green好題
題目意思是
告訴你要選出n個人（n為奇數(shù)）和總?cè)藬?shù) 和能提供的最大的幫助f
再告訴你每個人的成績和所需的aid，
然后我們要從其中找出n個人來，保證aid的和<=f的條件下，使得他們的中位數(shù)最大
這題乍一看摸不著頭腦，我們可以來分析一下
n為什么是奇數(shù)而不是偶數(shù)呢，顯然，奇數(shù)的話，中位數(shù)必然是一個固定的數(shù)，而不是兩個數(shù)的average
這樣想，我們有一點思路了
我們可以枚舉這個中位數(shù)，然后去驗證有沒有情況滿足
但是怎么驗證呢
首先，我們發(fā)現(xiàn)，有一部分必然不是中位數(shù)，這是前n/2小的和后n/2大的
所以我們先排一下序，只去枚舉中間的一段當中位數(shù)，假設(shè)當前枚舉第i個
那么我們必然要從左側(cè)選n/2個數(shù)，設(shè)其和為f1[i-1]，從右側(cè)選n/2個數(shù)，設(shè)其和為f2[i+1]
使得f1+f2+need[i]<=f，
我們用f1[i-1]表示左側(cè)中選出n/2個最小的，f2[i+1] 表示……
為什么和最小的呢，自己想去吧
然后就是怎么選呢，
這就用到最大堆的數(shù)據(jù)結(jié)構(gòu)
先考慮從左側(cè)選出n/2個使得和最小
我們維護一個元素個數(shù)為n/2的最大堆，
然后從n/2+1開始往堆中添加新元素，如果新元素小于堆頂，則添加并調(diào)整，
這樣，我們總是能保證選出的元素和的值最小
同樣右側(cè)選n/2個也是如此
這樣，這道題就解決了
最大堆維護最小和（dp）+枚舉
苦逼的看題啊，最后沒有結(jié)果輸出-1，我輸出0，wa了6次
哭……
code
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
#define maxn 200005
using namespace std;
struct node
{
int score,need;
}a[maxn],tmp1;
long long sum,tmp;
int nn;
int n,c,f;
int dp1[maxn],dp2[maxn];
struct heapnode
{
node x[maxn];
int num;
void nii(int n)
{
for(int i=1;i<=n/2;i++)
swap(x[i],x[n-i+1]);
}
long long getsum()
{
long long sum=0;
for(int i=1;i<=num;i++)
{
sum+=x[i].need;
}
return sum;
}
void down(int i,int m)
{
int t=2*i;
while(t<=m)
{
if(t<m&&x[t].need<x[t+1].need) t++;
if(x[i].need<x[t].need)
{
swap(x[i],x[t]);
i=t;
t=i*2;
}
else break;
}
}
void change(node tmpx)
{
x[1]=tmpx;
down(1,num);
}
void build()
{
for(int i=num/2;i>=1;i--) down(i,num);
}
}heap1,heap2;
bool cmp(node t1,node t2)
{
if(t1.score>t2.score)
return 0;
else if(t1.score<t2.score)
return 1;
else return t1.need<t2.need;
}
/**//*void print(heapnode t1)
{
printf("\n");
for(int i=1;i<=c;i++)
printf("%d %d\n",t1.x[i].score,t1.x[i].need);
printf("\n");
}*/
int main()
{
scanf("%d%d%d",&n,&c,&f);
for(int i=1;i<=c;i++) scanf("%d%d",&a[i].score,&a[i].need);
sort(a+1,a+c+1,cmp);
memcpy(heap1.x,a,sizeof(heap1.x));
memcpy(heap2.x,a,sizeof(heap2.x));
heap2.nii(c);
nn=n/2;
heap1.num=nn;
heap2.num=nn;
heap1.build();
heap2.build();
memset(dp1,0,sizeof(dp1));
memset(dp2,0,sizeof(dp2));
dp1[nn]=heap1.getsum();
dp2[nn]=heap2.getsum();
for(int i=nn+1;i<=c-nn;i++)
{
if(heap1.x[i].need<heap1.x[1].need)
{
dp1[i]=dp1[i-1]-heap1.x[1].need+heap1.x[i].need;
heap1.change(heap1.x[i]);
}
else dp1[i]=dp1[i-1];
}
for(int i=nn+1;i<=c-nn;i++)
{
if(heap2.x[i].need<heap2.x[1].need)
{
dp2[i]=dp2[i-1]-heap2.x[1].need+heap2.x[i].need;
heap2.change(heap2.x[i]);
}
else dp2[i]=dp2[i-1];
}
bool flag;
flag=0;
for(int i=c-nn;i>=nn+1;i--)
{
if(a[i].need+dp1[i-1]+dp2[c-i]<=f)
{
printf("%d\n",a[i].score);
flag=1;
break;
}
}
if(flag==0)
printf("-1\n");
return 0;
}