解题报告 之 HDU5303 Delicious Apples

解题报告 之 HDU5303 Delicious Apples

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：一个给定长度为L的圈。你在0位置。

圈上有n个苹果树（给出位置），每棵树下有一定数量的苹果。你有一个能装K个苹果的框子。如今你要将全部苹果运到你所在的位置。问你最少走多少路？

分析：第一点非常easy想到，离起点左右非常近的苹果更倾向于选择原路返回更省。而靠近中间的苹果有可能打包绕一圈再回来更省。

那么问题来了？这个“中间”究竟如何才算中间呢，这个点就是这个题的精髓。

大家能够把这个中间想成一个区间，问题的难点就转化为了怎么选择这个区间。也就是我们要决定选择哪些苹果绕一圈打包回来更省。

这的确是一个难解决的问题。。也是本题巧妙的地方。我们枚举究竟要把那些苹果绕一圈打包带回。我们把一个一个苹果离散出来而不再以苹果树作为考虑对象。并把这些苹果划分为更靠近左边和更靠近右边的。那么我们要枚举的就是，要从左边拿多少个（无疑是最远的那几个）到绕一圈的这个框中。

注意重点在于，从左边最多取K个，也就是仅仅有最后一筐可能须要绕一圈。由于假设从左边取超过K个。那么我们全然能够先取K个依照原路返回的方法（路程一定<=L），那之后问题终于回归不超过K个。

我们枚举的实质事实上在尝试一种平衡局面，使得左右两边选择原路返回路线的框能够尽量利用充分，而不要出现有一趟原路返回仅仅装1、2个苹果。

上代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>

using namespace std;
typedef long long ll;

const int MAXN = 1e5 + 10;
ll loc[MAXN];
ll disl[MAXN], disr[MAXN];//依照原路返回策略取完第i个苹果所走的路程
ll ans, L, n, k, cnt;
vector<ll> lloc, rloc; //存储从左右数的苹果位置

int main()
{
int kase;
scanf( "%d", &kase );
while(kase--)
{
memset( disl, 0, sizeof disl );
memset( disr, 0, sizeof disr );
lloc.clear();
rloc.clear();
cnt = 0;

scanf( "%lld%lld%lld", &L, &n, &k );
for(int i = 1; i <= n; i++)
{
ll location, number;
scanf( "%lld%lld", &location, &number );
for(int j = 1; j <= number; j++) //离散化
loc[cnt++] = location;

}

for(int i = 0; i < cnt; i++)
{
if(2 * loc[i] < L)
lloc.push_back( loc[i] );
else
rloc.push_back( L - loc[i] );
}//苹果分边

sort( lloc.begin(), lloc.end() );
sort( rloc.begin(), rloc.end() );

int left = lloc.size(), right = rloc.size();

for(int i = 0; i < left; i++)
disl[i + 1] = (i + 1 <= k ? lloc[i] : disl[i + 1 - k] + lloc[i]);
for(int i = 0; i < right; i++)
disr[i + 1] = (i + 1 <= k ? rloc[i] : disr[i + 1 - k] + rloc[i]);

ans = (disl[left] + disr[right]) * 2; //不绕一圈的情况，不要忘了

for(int i = 0; i <= left&&i <= k; i++)//枚举绕一圈的那框从左边取多少
{
ll pickl = left - i;
ll pickr = right - (k - i);

ans = min( ans, 2 * (disl[pickl] + disr[pickr]) + L );
}

printf( "%lld\n", ans );
}
return 0;
}