codeforces 251 div2 D. Devu and his Brother 三分

D. Devu and his Brother

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Devu and his brother love each other a lot. As they are super geeks, they only like to play with arrays. They are given two arrays a and bby their father. The array a is given to Devu and b to his brother.

As Devu is really a naughty kid, he wants the minimum value of his array a should be at least as much as the maximum value of his brother's array b.

Now you have to help Devu in achieving this condition. You can perform multiple operations on the arrays. In a single operation, you are allowed to decrease or increase any element of any of the arrays by 1. Note that you are allowed to apply the operation on any index of the array multiple times.

You need to find minimum number of operations required to satisfy Devu's condition so that the brothers can play peacefully without fighting.

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105). The second line will contain n space-separated integers representing content of the array a (1 ≤ ai ≤ 109). The third line will contain m space-separated integers representing content of the array b (1 ≤ bi ≤ 109).

Output

You need to output a single integer representing the minimum number of operations needed to satisfy Devu's condition.

题意：两个序列，每次只能加一减一，使得第一个序列的最小值要大于等于第二个序列的最大值，求最少的操作次数；

思路：一个凹性函数，利用三分求答案；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF ) return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
ll a[100010];
ll b[100010];
ll check(ll x,ll y,ll cha)
{
    ll sum=0;
    for(ll i=0;i<x;i++)
    if(cha>a[i])
    sum+=(cha-a[i]);
    for(ll i=0;i<y;i++)
    if(cha<b[i])
    sum+=b[i]-cha;
    return sum;
}
int main()
{
    ll x,y,z,i,t;
    scanf("%I64d%I64d",&x,&y);
    ll maxx=0;
    ll minn=1000000010;
    for(i=0;i<x;i++)
    {
        scanf("%I64d",&a[i]);
        minn=min(minn,a[i]);
    }
    for(i=0;i<y;i++)
    {
        scanf("%I64d",&b[i]);
        maxx=max(maxx,b[i]);
    }
    if(minn>=maxx)
    {
        printf("0
");
        return 0;
    }
    ll st=minn;
    ll en=maxx;
    ll mid,midd;
    ll ans=min(check(x,y,minn),check(x,y,maxx));
    while(st<en)
    {
        mid=(st+en)/2;
        midd=(mid+en)/2;
        ll  checkmid=check(x,y,mid);
        ll checkmidd=check(x,y,midd);
        if(checkmid>checkmidd)
        st=mid;
        else
        en=midd;
        ans=min(ans,min(checkmid,checkmidd));
    }
    printf("%I64d
",ans);
    return 0;
}