HDU 5303(Delicious Apples- 环上折半dp+贪心)

Delicious Apples

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Total Submission(s): 1585 Accepted Submission(s): 514



[align=left]Problem Description[/align]
There are n
apple trees planted along a cyclic road, which is
L
metres long. Your storehouse is built at position
0
on that cyclic road.

The ith
tree is planted at position xi,
clockwise from position 0.
There are ai
delicious apple(s) on the ith
tree.

You only have a basket which can contain at most K
apple(s). You are to start from your storehouse, pick all the apples and carry them back to your storehouse using your basket. What is your minimum distance travelled?

1≤n,k≤105,ai≥1,a1+a2+...+an≤105

1≤L≤109

0≤x[i]≤L

There are less than 20 huge testcases, and less than 500 small testcases.

[align=left]Input[/align]
First line: t,
the number of testcases.

Then t
testcases follow. In each testcase:

First line contains three integers, L,n,K.

Next n
lines, each line contains xi,ai.

[align=left]Output[/align]
Output total distance in a line for each testcase.

[align=left]Sample Input[/align]

2
10 3 2
2 2
8 2
5 1
10 4 1
2 2
8 2
5 1
0 10000

[align=left]Sample Output[/align]

18
26

[align=left]Author[/align]
XJZX

[align=left]Source[/align]
2015 Multi-University Training Contest 2

[align=left]Recommend[/align]
wange2014

半环拆成1半，左边的尽可能左走，右边的右走，

现在考虑一下绕圈：

由于2次绕圈可以拆成左右各来回一次，一定不劣，

所以考虑1次绕圈，显然取k个苹果最优

剩下的苹果由dp贪心得到。

dp[i]=dp[i-k] +pos[i] //显然第i次是去pos[i],相当于尽量取远的，剩下的就是dp[i-k]

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
ll dpl[MAXN],dpr[MAXN],L;
int n,K,n1,n2;
ll al[MAXN],ar[MAXN];
void work(ll *dp,int n,ll *pos)
{
sort(pos+1,pos+1+n);
For(i,n) {
if (i<=K) dp[i]=pos[i];
else dp[i]=pos[i]+dp[i-K];
}
}
int main()
{
//	freopen("D.in","r",stdin);
//	freopen(".out","w",stdout);

int T;cin>>T;
while(T--) {
cin>>L>>n>>K;
n1=n2=0;
MEM(al) MEM(ar) MEM(dpl) MEM(dpr)
For(i,n)
{
int xi,ai;
scanf("%d%d",&xi,&ai);
if (xi<=L/2) while (ai--) al[++n1]=xi;
else while (ai--) ar[++n2]=L-xi;
}
work(dpl,n1,al);
work(dpr,n2,ar);

ll ans=(dpl[n1]+dpr[n2])*2;

Rep(i,K+1) ans=min( 2LL*( dpl[max(n1-i,0)] + dpr[max(n2-(K-i),0)]  ) + L , ans  );
cout<<ans<<endl;
}
return 0;
}