Delicious Apples

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

Problem Description

There are  apple
trees planted along a cyclic road, which is  metres
long. Your storehouse is built at position  on
that cyclic road.

The th
tree is planted at position ,
clockwise from position .
There are  delicious
apple(s) on the th
tree.

You only have a basket which can contain at most  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?

There are less than 20 huge testcases, and less than 500 small testcases.

Input

First line: ,
the number of testcases.

Then  testcases
follow. In each testcase:

First line contains three integers, .

Next  lines,
each line contains .

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

解题思路:

注意到，最多仅仅有一次会绕整个圈走一次。因此，先贪心的处理左半环和右半环。然后枚举绕整圈的时候从左側摘得苹果和从右側摘得苹果的数目。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
const int MAXN = 100000 + 10;
int L, N, K;
LL x[MAXN];
LL ld[MAXN], rd[MAXN];
vector<LL>l, r;
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%d", &L, &N, &K);
        l.clear(); r.clear();
        int pos, num, m = 0;
        for(int i=1;i<=N;i++)
        {
            scanf("%d%d", &pos, &num);
            for(int i=1;i<=num;i++)
                x[++m] = (LL)pos;
        }
        for(int i=1;i<=m;i++)
        {
            if(2 * x[i] < L) l.push_back(x[i]);
            else r.push_back(L - x[i]);
        }
        sort(l.begin(), l.end()); sort(r.begin(), r.end());
        int lsz = l.size(), rsz = r.size();
        memset(ld, 0, sizeof(ld)); memset(rd, 0, sizeof(rd));
        for(int i=0;i<lsz;i++)
            ld[i + 1] = (i + 1 <= K ? l[i] : ld[i + 1 - K] + l[i]);
        for(int i=0;i<rsz;i++)
            rd[i + 1] = (i + 1 <= K ? r[i] : rd[i + 1 - K] + r[i]);
        LL ans = (ld[lsz] + rd[rsz]) * 2;
        for(int i=0;i<=lsz&&i<=K;i++)
        {
            int p1 = lsz - i;
            int p2 = max(0, rsz-(K-i));
            ans = min(ans, 2*(ld[p1] + rd[p2]) + L);
        }
        cout << ans << endl;
    }
    return 0;
}