HDU 5303 Delicious Apples（贪心 + 背包 2015多校啊）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=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

Source

2015 Multi-University Training Contest 2

题意：

一个长为 L 的环行路线。有 n 颗苹果树，每颗苹果树上有 a[i] 个苹果，一个人在0点（仓库的位置），他有一个篮子，篮子每次最多仅仅能装 k 个苹果。求要装全然部的苹果而且回到仓库的最小路程；给出的苹果树坐标是按顺时针的。

官方题解：



PS：

贪心。把环从中间分为两段。分左右两条线。

利用 a[i] 数组记录每一个苹果所在的苹果树的位置，之后再将苹果依照所在的位置进行排序一下。

所以我们就知道了每次摘 k 个苹果的路程是最远的那个苹果所在的位置。

再用 sum[i] 表示摘第 i 个苹果时的最小代价和。

依据背包的思想得到：

if ( i <= k )

sum[i] = d[i]

else

sum[i] = d[i] + sum[i-k]
注意：

另一种情况是在最后当剩下的苹果少于等于 k 个时，也许一次性绕环一圈拿完最后的k个所需的路程更少。

枚举剩下的最后k个！

代码例如以下：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define LL __int64
const int maxn = 100047;
vector <int> v1, v2;
LL a[maxn];
LL sum1[maxn], sum2[maxn];

int main()
{
int t;
int n, k, l;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&l,&n,&k);
int pos, num;
int h = 0;
for(int i = 0; i < n; i++)
{
scanf("%d%d",&pos,&num);
for(int j = 0; j < num; j++)
{
a[++h] = pos;
}
}
k = min(h,k);
v1.clear();
v2.clear();
for(int i = 1; i <= h; i++)
{
if(a[i]*2 < l)
{
v1.push_back(a[i]);
}
else
{
v2.push_back(l - a[i]);
}
}
memset(sum1,0,sizeof(sum1));
memset(sum2,0,sizeof(sum2));
sort(v1.begin(), v1.end());
sort(v2.begin(), v2.end());
int len1 = v1.size(), len2 = v2.size();
for(int i = 0; i < len1; i++)
{
int id = i+1;
if(id <= k)
{
sum1[id] = v1[i];
}
else
{
sum1[id] = v1[i]+sum1[id-k];
}
}
for(int i = 0; i < len2; i++)
{
int id = i+1;
if(id <= k)
{
sum2[id] = v2[i];
}
else
{
sum2[id] = v2[i]+sum2[id-k];
}
}

LL ans = 2*(sum1[len1]+sum2[len2]);//来回
int t1, t2;
for(int i = 0; i <= k && i <= len1; i++)
{
t1 = len1 - i;
t2 = len2-(k-i);
if(t2 < 0)
{
t2 = 0;
}
ans = min(ans,2*(sum1[t1]+sum2[t2])+l);//最后不足k个绕行一圈所有摘走
}
printf("%I64d\n",ans);
}
return 0;
}