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Moo University - Financial Aid
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 7491 | Accepted: 2159 |
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.
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
* 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
这道题直觉要使用二分法，但需要证明单调性。
设区间【L ，R）的中点是M，对其左右分别排序，掐出前n/2个最小的数，左面那n/2个数的和为left,右面的n/2的两个数的和为right，如果left,right，本身这三个和相加大于了f，如果区间向右移动，只能是left不变，right 更大，所以得向左移动，这样就证明了单调性。
但是它要求中位数，如果求完了，mid的左侧或右侧不足n/2个数，那么就说明没有对应的方案。
代码如下：
#include
#include
#include
#include
using namespace std;
int n,c,f,s;
struct node
{
int c,f;//CÊÇ¿¼ÊÔ·ÖÊý,fÊDz¹Öú
bool operator < (const node &a) const
{
return this->c < a.c;
}
}cow[100010];
bool isok(int mid)
{
int temp[100010];
for(int i=1;i<=c;i++)
temp[i]=cow[i].f;
sort(temp+1,temp+mid);
sort(temp+mid+1,temp+c+1);
int left=0,right=0;
for(int i=1;i<=s;i++)
{
left+=temp[i];
right+=temp[mid+i];
}
if(left+right+cow[mid].f<=f) return true;
else return false;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("finance.in","r",stdin);
#endif // ONLINE_JUDGE
scanf("%d %d %d",&n,&c,&f);
s=n/2;
for(int i=1;i<=c;i++)
scanf("%d %d",&cow[i].c,&cow[i].f);
sort(cow+1,cow+c+1);
int l=1,r=c+1;
while(r>l)
{
int mid=(l+r)/2;
if(c-mid