Hdu 5303 Delicious Apples 2015 Multi-University Training Contest 2

Delicious Apples

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

Total Submission(s): 782    Accepted Submission(s): 263


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

2015 Multi-University Training Contest 2

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define sf scanf
#define mx 2111111
#define LL long long
LL n,m,k;
LL d[mx],d1[mx],d2[mx],f1[mx],f2[mx];
int main()
{

LL T;
sf("%lld",&T);
while(T--)
{
sf("%lld%lld%lld",&n,&m,&k);
long long cnt = 0;
for(LL i = 0;i<m;i++)
{
long long x,y;
sf("%lld%lld",&x,&y);
while(y){d[cnt++] = x;y--;}
}
m= cnt;
k = min(k,m);
long long n1=0,n2=0;
for(LL i = 0;i<m;i++)
if(d[i]*2<=n) d1[++n1] = d[i];else d2[++n2] = n-d[i];
//cout<<n1<<"--"<<n2<<endl;
sort(d1+1,d1+n1+1);
sort(d2+1,d2+n2+1);
for(LL i = 1;i<=n1;i++)
if(i<=k)f1[i] =d1[i];
else f1[i] = f1[i-k]+d1[i];
//   for(LL i =0;i<=n1;i++)
//     cout<<i<<":"<<f1[i]<<" --- "<<d1[i]<<endl;
for(LL i=1;i<=n2;i++)
if(i<=k)f2[i] = d2[i];
else f2[i] = f2[i-k] +d2[i];
// cout<<f1[n1-1] <<" "<<f2[n2-1]<<endl;
long long ans = 0;
if(n1>0) ans+= f1[n1];
if(n2>0) ans+= f2[n2];
ans*=2;
//cout<<ans<<endl;
for(LL i = max(0LL,n1-k);i<n1;i++)
{
LL j = max(0LL,n2-(k-(n1-i)));
//     cout<<i<<" "<<j<<endl;
//   cout<<f1[i]<<" "<<f2[j]<<endl;
ans = min(ans,(f1[i]+f2[j])*2+n);
}
printf("%lld\n",ans);
}
}