Delicious Apples (hdu 5303 贪心+枚举)

Delicious Apples

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

Total Submission(s): 395    Accepted Submission(s): 122

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

 

Source

field=problem&key=2015+Multi-University+Training+Contest+2&source=1&searchmode=source">2015 Multi-University Training Contest 2

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  

pid=5309" target="_blank">5309 5308 5307 5306 5305 

题意：在一个圆上有n个苹果树。告诉苹果树的位置和每棵树上的苹果个数，另一个容量为K的篮子，用篮子去摘苹果，起点在位置0，重复去摘直到把全部的苹果都摘回到0，问走的最短距离为多少。

思路：首先将圆一分为二，在圆形两側能拿满的话肯定就是仅仅走半边再回去，这样比走整圈划算。另外还要想到最后两边都不足K个了。这个时候最多须要走一个整圈，我们不知道这个整圈拿了哪几个苹果，那么就枚举K个。

比赛时仅仅是想到了贪心，最后那一部分没有枚举，另外这里的苹果进行了离散化，由于苹果总数仅仅有1e5，大大简化了代码。自己当时写的太冗余=-=

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1e5+10;
const int MAXN = 2005;
const int N = 1005;

ll Len,n,k,cnt;
ll pos[maxn],dp_l[maxn],dp_r[maxn]; //dp[i]表示拿完前i个苹果要走的的距离和
vector<ll>posl,posr;

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    ll i,j,t;
    scanf("%lld",&t);
    ll x,num;
    while (t--)
    {
        cnt=1;
        scanf("%lld%lld%lld",&Len,&n,&k);
        posl.clear();
        posr.clear();
        for (i=0;i<n;i++)
        {
            scanf("%lld%lld",&x,&num);
            for (j=0;j<num;j++)
                pos[cnt++]=x;
        }
        cnt--;
        k=min(k,cnt);
        for (i=1;i<=cnt;i++)
        {
            if (2*pos[i]<Len)
                posr.push_back(pos[i]);
            else
                posl.push_back(Len-pos[i]);
        }
        sort(posl.begin(),posl.end());
        sort(posr.begin(),posr.end());
        dp_l[0]=dp_r[0]=0;
        for (i=0;i<(ll)posl.size();i++)
        {
            if (i+1<=k)
                dp_l[i+1]=posl[i];
            else
                dp_l[i+1]=dp_l[i+1-k]+posl[i];
        }
        for (i=0;i<(ll)posr.size();i++)
        {
            if (i+1<=k)
                dp_r[i+1]=posr[i];
            else
                dp_r[i+1]=dp_r[i+1-k]+posr[i];
        }
        ll ans=(dp_l[posl.size()]+dp_r[posr.size()])*2;
        for (i=0;i<=posr.size()&&i<=k;i++)
        {
            ll right=posr.size()-i;
            ll left=max((ll)0,(ll)(posl.size()-(k-i)));
            ans=min(ans,(ll)Len+(dp_l[left]+dp_r[right])*2);
        }
        printf("%lld\n",ans);
    }
    return 0;
}