hdu 5303 Delicious Apples 2015多校联合训练赛2 dp+枚举

Delicious Apples

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 311    Accepted Submission(s): 92

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  i th tree is planted at position  xi , clockwise from position  0 . There are  ai  delicious apple(s) on the  i th 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

2015 Multi-University Training Contest 2

题意：

有一个圈，圈上不同位置有一定数量的苹果，苹果与起点有个距离。初始位置在0点。有个篮子，每次可以装K个苹果。问，要把所有的苹果都用篮子装到起点，至少要走多远的距离。

分析：

苹果总数 <= 100000. 那么按顺序排序每个苹果。

按顺时针，和逆时针分别处理。

dp[0][i]表示只能顺时针向前走，逆时针回到起点。把顺时针的前i个苹果带回原点的最小路程

            那么dp[0][i] = dp[0][i-K] + 2*dist[i]   -----------dist[i]表示第i个苹果顺时针距离起点的距离

同理逆时针的转移方程为

           dp[1][i] = dp[1][i+K] + 2*(L-dist[i])

然后枚举第i个苹果是分界线，i个苹果和之前的苹果是顺时针取得，i+1和之后的苹果是逆时针取得

         那么ans = min(ans,dp[0][i]+dp[1][i+1])

但是存在这样的情况，取走第i个苹果的时候，不原路放回，二是直接向前走到终点。

       那么ans = min(ans,dp[0][i-K]+dp[1][i+1] + L)

最后ans就是答案

  

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define maxn 100007
struct Point{
    int x;
    int num;
};
Point p[100007];
Point px[maxn];
int cmp(Point a,Point b){
    return a.x < b.x;
}
ll dp[2][maxn];
int main(){
    int t,L,n,K;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&L,&n,&K);
        int cnt = 1,x,y;
        for(int i = 0;i < n; i++){
            scanf("%d%d",&px[i].x,&px[i].num);
        }
        sort(px,px+n,cmp);
        for(int i = 0;i < n; i++){
            while(px[i].num--){
                p[cnt++].x = px[i].x;
            }
        }
        memset(dp[0],0,sizeof(ll)*(cnt+3));
        memset(dp[1],0,sizeof(ll)*(cnt+3));
        for(int i = 1;i < cnt; i++){
            dp[0][i] = dp[0][max(0,i-K)] + p[i].x * 2;
        }
        for(int i = cnt-1;i > 0; i--){
            dp[1][i] = dp[1][min(cnt,i+K)] + (L - p[i].x) * 2;
        }
        ll ans = min(dp[0][cnt-1],dp[1][1]);
        for(int i = 1;i < cnt; i++){
            ans = min(ans,dp[0][i]+dp[1][i+1]);
            ans = min(ans,dp[0][max(0,i-K)] + dp[1][i+1] + L);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}