HDU 5303 Delicious Apples

Delicious Apples

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

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.

 

Input

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.

 

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

 刚看这个题的时候头都大了，感觉贪心起来好复杂啊。情况太多，不是很好处理。但是多列几种情况就会发现，绕圈的话最多只需要绕一次，而且还必须是左半圈的苹果数加上右半圈的苹果数小于K时才有可能需要绕圈（自己举两个例子就明白了）。所以对所有需要绕圈情况枚举判断看看是不是更优即可。因为苹果总数不超过1e5，所以可以将苹果离散化，这样更好处理。用dpl[]和dpr[]分别表示取到某个苹果所需要的距离，这里需要用到贪心，想一想再不考虑其他情况下怎么取苹果才最优。例如两个苹果树1  k+1和2 k， 明显是要优先将远处的苹果取为一篮子，依据这个规则更新dp数组即可。代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define maxn 100010
#define ll long long
using namespace std;
long long dpl[maxn],dpr[maxn],dis[maxn];
vector<int> vl,vr;
int main()
{
        ios::sync_with_stdio(false);
        int t;
        cin>>t;
        while(t--)
        {
                int l,n,k,cnt=1;
                vl.clear();
                vr.clear();
                cin>>l>>n>>k;
                for(int i=0;i<n;i++)
                {
                        int a,b;
                        cin>>a>>b;
                        for(int i=0;i<b;i++)
                        {
                                dis[cnt++]=a;
                                if(a<=(l/2))
                                {
                                        vl.push_back(a);
                                }
                                else
                                {
                                        vr.push_back(l-a);
                                }
                        }
                }
                sort(vl.begin(),vl.end());
                sort(vr.begin(),vr.end());
                dpl[0]=0;
                dpr[0]=0;
                for(int i=0;i<vl.size();i++)
                {
                        if(i+1<=k)
                        {
                                dpl[i+1]=vl[i];
                        }
                        else
                        {
                                dpl[i+1]=vl[i]+dpl[i+1-k];
                        }
                }
                for(int i=0;i<vr.size();i++)
                {
                        if(i+1<=k)
                        {
                                dpr[i+1]=vr[i];
                        }
                        else
                        {
                                dpr[i+1]=vr[i]+dpr[i+1-k];
                        }
                }
                long long ans=dpl[vl.size()]*2+dpr[vr.size()]*2;
                for(int i=0;i<=k&&i<=vl.size();i++)
                {
                        if(dpl[vl.size()-i]*2+dpr[max(0,(int)vr.size()-(k-i))]*2+l<ans)
                        {
                                ans=dpl[vl.size()-i]*2+dpr[max(0,(int)vr.size()-(k-i))]*2+l;
                        }
                }
                cout<<ans<<endl;
        }
        return 0;
}