2015 多校赛 第二场 1004 hdu(5303)

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

 

题意：给一个圈，设起点为0，给出周长为L，在圈中有 n 棵树，依次给出按顺时针方向与起点的距离，树上的苹果数。给出篮子大小（最多能装的苹果树）。问最少走多远可以采集完所有苹果。

 

思路：

以过起点的直径为界，以苹果树在左半圆和右半圆分别贪心，得到要走的距离。最后枚举最后走一整圈所采集的苹果（详见代码）。比较得出最小值。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 100050  
int d_l[maxn],d_r[maxn],l,n,k,t,x,a,L,R;
long long tot_l[maxn],tot_r[maxn],ans,tmp;
int main(){
    scanf("%d",&t);
    while(t--){
        L=R=0;
        scanf("%d%d%d",&l,&n,&k);
        for(int i=0;i<n;i++){
            scanf("%d%d",&x,&a);
            for(int j=0;j<a;j++){
                if(x*2<l) d_l[++L]=x;
                else d_r[++R]=l-x;
            }
        }
        sort(d_l+1,d_l+L+1);
        sort(d_r+1,d_r+R+1);
        for(int i=1;i<=L;i++){
            if(i<=k) tot_l[i]=d_l[i];
            else tot_l[i]=tot_l[i-k]+d_l[i];
        }
        for(int i=1;i<=R;i++){
            if(i<=k) tot_r[i]=d_r[i];
            else tot_r[i]=tot_r[i-k]+d_r[i];
        }
        ans=(tot_l[L]+tot_r[R])*2;
        for(int i=0;i<=k;i++){
            tmp=(tot_l[L-i]+tot_r[max(0,R-(k-i))])*2;
            ans=min(ans,l+tmp);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}