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.

First line: , the number of testcases.
Then  testcases follow. In each testcase:
First line contains three integers, .
Next  lines, each line contains .

Output total distance in a line for each testcase.

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

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;
}