poj 2010 Moo University - Financial Aid 最大化中位数 二分搜索 以后需要慢慢体会
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 6599 | Accepted: 1926 |
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
Source
#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
struct Node{
int s,m,id;
}node[20005];
int n,c,f,k;
bool cmp(Node a,Node b)
{
return a.m<b.m;
}
int ok(int mid)
{
int cnt1=0,cnt2=0,total=0;
for(int i=1;i<=c;i++)
if(node[i].s<mid&&cnt1<k)
{
cnt1++;
total+=node[i].m;
} //在加进一头牛时没有判断假设加进后总费用是否<=f;
else if(node[i].s>mid&&cnt2<(n-k))
{
cnt2++;
total+=node[i].m;
}
if(cnt1<k||cnt2<(n-k)||total>f)
return 0;
else return 1;
}
int main()
{
while(~scanf("%d %d %d",&n,&c,&f))
{
int maxn=0;
for(int i=1;i<=c;i++)
{
scanf("%d %d",&node[i].s,&node[i].m);
if(maxn<node[i].s)
maxn=node[i].s;
}
sort(node+1,node+1+c,cmp);
int l=0,r=maxn+1;
k=(n+1)/2;
while(r-l>1)
{
int mid=(l+r)/2;
if(ok(mid))
r=mid;
else
l=mid;
}
printf("%d\n",r-1);
}
return 0;
}
第二份是AC代码:总的思想是先进行score排序,记录好数据后,二分枚举可能的取到的中位(因为score已经排好序,所以尽量往id大的取),然后money排序,进行贪心的选取。
#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
struct Node{
int s,m,id;
}node[100005];
int n,c,f,k,score[100005],weight[100005];
bool cmpm(Node a,Node b)
{
return a.m<b.m;
}
bool cmps(Node a,Node b)
{
return a.s<b.s;
}
int main()
{
while(~scanf("%d %d %d",&n,&c,&f))
{
for(int i=1;i<=c;i++)
scanf("%d %d",&node[i].s,&node[i].m);
sort(node+1,node+1+c,cmps);
for(int i=1;i<=c;i++)
{
node[i].id=i;
score[i]=node[i].s;
weight[i]=node[i].m;
}
int ans=-1,l=0,r=c+1;
sort(node+1,node+1+c,cmpm);
k=n/2;
while(r-l>1)
{
int mid=(l+r)>>1;
int total=weight[mid],cnt1=0,cnt2=0;
for(int i=1;i<=c;i++)
if(cnt1<k&&node[i].id<mid&&node[i].m+total<=f)
{
cnt1++;
total+=node[i].m;
}
else if(cnt2<k&&node[i].id>mid&&node[i].m+total<=f)
{
cnt2++;
total+=node[i].m;
}
if(cnt1<k&&cnt2<k)
break;
else if(cnt1>=k&&cnt2>=k)
{
l=mid;//尽量往右选取
ans=score[l];
}
else if(cnt1<k)
l=mid;
else if(cnt2<k)
r=mid;
}
printf("%d\n",ans);
}
return 0;
}
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