HDU 5303 Delicious Apples (2015多校第二场贪心 + 枚举)-优快云博客

解决一个关于沿环形路径摘取苹果并返回仓库的问题，通过贪心算法与枚举策略结合，实现最小路径长度的计算。

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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

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

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;
}