cf343c 二分答案+模拟

/*
怎么判断能否在时间k内完成扫描
贪心：每次取出最靠左边的磁头去扫描最左边的，然后再往右扫描即可
如果当前点无法扫到最左侧点，那么后继点一样无法扫到
*/
#include<bits/stdc++.h>
#define maxn 100005
#define ll long long
using namespace std;

int n,m;
ll h[maxn],p[maxn];

int judge(ll x){
    ll time1,time2;
    int index=;
    for(int i=;i<=n;i++){
        if(h[i]-p[index]>x) return ;
        if(p[index]>=h[i]){//直接往右扫描
            while(index<=m && p[index]<=x+h[i])
                index++;
            if(index>m) return ;//
        }
        else {
            time1=(x-(h[i]-p[index]))/;
            time2=x-(h[i]-p[index])*;
            time1=max(time1,time2); //先往左在往右或者先往右再往左的最长右走时间
            while(index<=m && p[index]<=time1+h[i])
                index++;
            if(index>m) return ;
        }
    }
    return ;
}

int main(){
    while(scanf("%d%d",&n,&m)==){
        for(int i=;i<=n;i++) scanf("%lld",&h[i]);
        for(int i=;i<=m;i++) scanf("%lld",&p[i]);
        ll l=,r=max(h[n],p[m])*,ans=;
        while(l<=r){
            ll mid=l+r>>;
            if(judge(mid))
                ans=mid,r=mid-;
            else l=mid+;
        }
        printf("%I64d\n",ans);
    }
}