HDU 5303 Delicious Apples (贪心 枚举 好题)

Delicious Apples

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Total Submission(s): 199 Accepted Submission(s): 54

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.

[align=left]Input[/align]

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.

[align=left]Output[/align]

Output total distance in a line for each testcase.

[align=left]Sample Input[/align]

2
10 3 2
2 2
8 2
5 1
10 4 1
2 2
8 2
5 1
0 10000

[align=left]Sample Output[/align]

18
26

[align=left]Source[/align]

2015 Multi-University Training Contest 2

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5303

题目大意：有一个长为L的环。n个苹果树，一个篮子最多装k个苹果，装完要回到起点卸下再出发。给出n个苹果树顺时针的位置及苹果的个数。求摘全然部苹果走的最小路程

题目分析：显然。仅仅有在某种特殊条件下，即两側都还有苹果且能够一次装完且最后的苹果都离起点比較远。这样的情况下。我们直接绕圈可能会更优，也就是说整圈最多绕一次。因此我们能够先对两边贪心。题目的数据显示苹果的数量最多就1e5，显然我们能够把苹果“离散”出来，用x[i]记录第i个苹果到起点的位置。然后对位置从小到大排序，先选择路程小的。选择的时候用dis[i]记录单側装了i个苹果的最小路程，类似背包计数的原理。答案要乘2，由于是来回的，最后在k>=i时。枚举绕整圈的情况，szl－i表示仅仅走左边採的苹果数，szr -
(k - i)表示仅仅走右边採的苹果树。画个图就能看出来了。注意右边这里可能值为负。要和0取最大，然后答案就是（disl[szl－i] + disr[szr - (k - i)]）* 2 ＋L，这里事实上绘图更加直观。最后取最小就可以，注意有几个wa点，一个是要用long long。二是之前说的出现负数和0取大，三是每次要清零

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 1e5 + 5;
int  L, n, k;
ll x[MAX], disl[MAX], disr[MAX];
vector <ll> l, r;

int main()
{
int T;
scanf("%d", &T);
while(T--)
{
memset(disl, 0, sizeof(disl));
memset(disr, 0, sizeof(disr));
l.clear();
r.clear();
scanf("%d %d %d", &L, &n, &k);
int cnt = 1;
for(int i = 1; i <= n; i++)
{
ll pos, num;
scanf("%lld %lld", &pos, &num);
for(int j = 1; j <= num; j++)
x[cnt ++] = (ll) pos;   //离散操作
}
cnt --;
for(int i = 1; i <= cnt; i++)
{
if(2 * x[i] < L)
l.push_back(x[i]);
else
r.push_back(L - x[i]);  //记录位置
}
sort(l.begin(), l.end());
sort(r.begin(), r.end());
int szl = l.size(), szr = r.size();
for(int i = 0; i < szl; i++)
disl[i + 1] = (i + 1 <= k ? l[i] : disl[i + 1 - k] + l[i]);
for(int i = 0; i < szr; i++)
disr[i + 1] = (i + 1 <= k ? r[i] : disr[i + 1 - k] + r[i]);
ll ans = (disl[szl] + disr[szr]) * 2;
for(int i = 0; i <= szl && i <= k; i++)
{
int p1 = szl - i;
int p2 = max(0, szr - (k - i));
ans = min(ans, 2 * (disl[p1] + disr[p2]) + L);
}
printf("%I64d\n", ans);
}
}