Delicious Apples(多校联合训练） 分类： ACM 多校 2015-07-24 17:02 18人阅读 评论(0) 收藏

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 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.

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

Source

2015 Multi-University Training Conest 2
这题若是道路不是圈的话，就左右两边贪心可得最短路程，但由于道路是圆圈，所以可以走整圈。具体的分析在代码。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<set>
#include<queue>
#include<set>
#include<vector>
#include<math.h>
#define LL long long
const int N=100000+50;
using namespace std;
int X[N],a[N],b[N];
LL sum1[N],sum2[N];
int main()
{
    int t,l,num,k;
    cin>>t;
    while(t--)
    {
        memset(a,0,sizeof(a)) ;
        memset(b,0,sizeof(b)) ;
        memset(sum1,0,sizeof(sum1)) ;
        memset(sum2,0,sizeof(sum2)) ;
        cin>>l>>num>>k;
        int x,y;
        int n=0;
        while(num--)
        {
            cin>>x>>y;
            for(int i=1;i<=y;i++)
            X[++n]=x;//先求出每个苹果的位置。 
        }
        int n1=0,n2=0; 
        for(int i=1;i<=n;i++)
        {
            if(2*X[i]<=l)
            a[++n1]=X[i];//原点右边的每个苹果所在的位置 
            else
            b[++n2]=l-X[i];//原点左边的每个苹果所在的位置 
        }
        sort(a+1,a+1+n1);
        sort(b+1,b+1+n2);
        sum1[0]=sum2[0]=0; 
        for(int i=1;i<=n1;i++)
        {
            if(i<=k)
            sum1[i]=a[i];
            else
            sum1[i]=sum1[i-k]+a[i]; 
        } //算出摘完右边的苹果所用的的路程 
        for(int i=1;i<=n2;i++)
        {
            if(i<=k)
            sum2[i]=b[i];   
            else
            sum2[i]=sum2[i-k]+b[i];
        }//算出摘完左边的苹果所用的的路程 
        LL ans=(sum1[n1]+sum2[n2])*2;
        for(int i=0;i<=n1&&i<=k;i++)//枚举若是摘右边的苹果时走整圈时的距离，然后和不走整圈时哪个答案小 
        {
            int left=n1-i;//右边所剩的苹果数 
            int right=max(0,n2-(k-i));//装完篮子后左边所剩的苹果数 
            ans=min(ans,l+(sum1[left]+sum2[right])*2);
        }
        printf("%lld\n",ans); 
    }
    return 0;
}

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