Delicious Apples
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2804 Accepted Submission(s): 878
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
题意：有一个周长为l的圈，你家在的地方为0点，顺时针标号1~l-1。有n(n<=1e5)颗苹果树，每棵树所在位置为x[i],每颗树上有苹果a[i]个，你有一个容量为k的篮子（一次最多装k个苹果），问最少路程为多少可以把所有苹果摘回家。
思路：采用dp的思想。注意到题目有a1+a2+...+an≤105 自然想到枚举每一个苹果从左边拿还是从右边拿。
于是先把每颗苹果树按x值从小到大排序，再依次把每个苹果的位置都存到数组里，然后预处理从右、左边直接拿这个苹果的需要走的路程。然后再枚举从第i+1个苹果开始从左边拿，取最小值。考虑到省路程的情况只有一种，就是走一圈。而且越拿靠近中间的苹果越好。于是再枚举一下第i+1个苹果到第i+k个苹果走一圈拿走，剩下的同上，取一个最小值。注意，如果苹果总数小于k我们是没有枚举到走一圈的情况的，所以再特判一下。
思路很好理解，但是真正到赛场上想出来还是很难。
代码：
#include<bits/stdc++.h>
#define ll long long
#define maxn 200010
#define inf 0x3f3f3f3f
using namespace std;
struct node
{
int x,v;
node(){}
node(int aa,int bb){x=aa;v=bb;}
bool operator<(node aa)const
{
return x<aa.x;
}
}a[maxn];
ll b[maxn];
int n,m,k,flag,f,g;
ll ans,tmp,cnt,l;
ll dpl[maxn],dpr[maxn];
int main()
{
int T,cas=1;
scanf("%d",&T);
while(T--)
{
scanf("%lld%d%d",&l,&n,&k);
ans=0; f=0;g=0;
memset(dpl,0,sizeof(dpl));
memset(dpr,0,sizeof(dpr));
for(int i=0;i<n;i++)
{
int xx,vv;
scanf("%d%d",&xx,&vv);
a[f++]=node(xx,vv);
}
sort(a,a+f);
g=1;
for(int i=0;i<n;i++)
for(int j=0;j<a[i].v;j++)
b[g++]=(ll)a[i].x;
for(int i=1;i<g;i++)
{
int j=max(0,i-k);
dpr[i]=dpr[j]+2*b[i];
}
for(int i=g-1;i>=1;i--)
{
int j=min(i+k,g);
dpl[i]=dpl[j]+2*(l-b[i]);
}
for(int i=0;i<g;i++)
{
if(ans==0) ans=dpr[i]+dpl[i+1];
else ans=min(ans,dpr[i]+dpl[i+1]);
}
for(int i=0;i+k<g;i++)
{
ans=min(ans,dpr[i]+dpl[i+1+k]+l);
}
if(g<=k) ans=min(ans,l);
printf("%lld\n",ans);
}
return 0;
}