hdu 5303 Delicious Apples(2015 Multi-University Training Contest 2)

Delicious Apples

                                                                            Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
                                                                                                      Total Submission(s): 315    Accepted Submission(s): 92

Problem Description

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

 

Input

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 .

 

Output

Output total distance in a line for each testcase.

 

Sample Input

   
   
   
   

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

 

Sample Output

   
   
   
   

    
    
    
    18
26
   
   
   
   

 

题目大意：

    在一条环形通路上，有n棵苹果树，每棵苹果树有ai个苹果，求最少的行走距离用容量为k的篮子将苹果都运回起点。

解题思路：

     比赛的时候想的方法跟题解一样，但是一直没想到怎么实现，先做的其他题，后来也没空做这道题了。在左右两边

的且可以装满整框的肯定运回离它最近的起点就可以了，还有可能越过中轴线才可以装满一筐，但是只有可能1次。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=200000+100;
long long dp1[maxn];
long long dp2[maxn];
int a[maxn],b[maxn];
int main()
{
    int n,k,l,t;
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&l,&n,&k);
        int cur1=0,cur2=0,u,v;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            for(int j=0;j<v;j++)
            {
                if(u*2<l)
                a[++cur1]=u;
                else
                b[++cur2]=l-u;
            }
        }
        sort(a+1,a+cur1+1);
        sort(b+1,b+cur2+1);
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        for(int i=1;i<=cur1;i++)
        {
            if(i<=k)
            dp1[i]=a[i];
            else
            dp1[i]=dp1[i-k]+a[i];
        }
        for(int i=1;i<=cur2;i++)
        {
            if(i<=k)
            dp2[i]=b[i];
            else
            dp2[i]=dp2[i-k]+b[i];
        }
        long long ans=(dp1[cur1]+dp2[cur2])*2;
        for(int i=0;i<=cur1&&i<=k;i++)
        {
            int le=cur1-i,re=max(0,cur2-(k-i));
            ans=min(ans,l+(dp1[le]+dp2[re])*2);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}