2015 多校联赛 ——HDU5303（贪心）

Delicious Apples

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1371    Accepted Submission(s): 448

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

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 2

题意：给定一个环，以下标为处为起始点，距离起始点Xi位置种植一颗苹果树，该树有a个苹果，篮子的最大容量为K，那么求摘完全部苹果所需的最短距离。

当时大致知道方向，但是没想到具体方法。首先贪心左右半边，推出len-k时的最短路程，最后则可以直接走一圈取完。然后再与左右分别取相比较，取较小。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAX 100050
long long sum_left[MAX];
long long sum_right[MAX];
int l[MAX],r[MAX];

long long ans;
int Left,Right;
int n,k,len,T,a,b;

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        memset(sum_left,0,sizeof(sum_left));
        memset(sum_right,0,sizeof(sum_right));

        scanf("%d%d%d",&len,&n,&k);
        Left=0,Right=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&a,&b);
            for(int j=0; j<b; j++)
            {
                if(a*2<len)
                   l[++Left]=a;
                else
                    r[++Right]=len-a;
            }
        }
        sort(l+1,l+Left+1);
        sort(r+1,r+Right+1);
        for(int i=1; i<=Left; i++)
        {
            if(i<=k)
                sum_left[i]=l[i];
            else
                sum_left[i]=sum_left[i-k]+l[i];
        }
        for(int i=1; i<=Right; i++)
        {
            if(i<=k)
                sum_right[i]=r[i];
            else
                sum_right[i]=sum_right[i-k]+r[i];
        }
        ans=(sum_left[Left]+sum_right[Right])*2;

        for(int i=0; i<=k && i <= Left; i++)
        {
            long long lll = (sum_left[Left-i]+sum_right[max(0,Right-(k-i))])*2;
            ans=min(ans,len+lll);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}