HDU 5303 Delicious Apples (2015多校第二场 贪心 + 枚举)

Delicious Apples

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

Total Submission(s): 321    Accepted Submission(s): 95

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 

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tree is planted at position 

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clockwise from position .
There are 

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apple(s) on the 

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tree.



You only have a basket which can contain at most 

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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?

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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, 

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Next 

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each line contains 

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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;
}