對每個人行道求出移動距離在哪些區間內時其在建築物前面。現在問題即為選一個點使得其被最多的區間包含。差分即可。對建築暴力去掉重疊部分。開始時沒有去重用了nm次vector的push_back，時間大概是去重寫法的300倍，不知所措。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace
 
 std;
#define ll long long
#define N 10010
#define M 1010
#define K 1000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    
 
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N],delta[K<<1],cnt,d=K,s;
struct data
{
    int l,r;
    bool operator <(const data&a) const
    {
        return
 
 l<a.l;
    }
}b[M],c[M];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4925.in","r",stdin);
    freopen("bzoj4925.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=m;i++) b[i].l=read(),b[i].r=read();
    for (int i=1;i<=m;i++)
    {
        bool flag=1;
        for (int j=1;j<=m;j++)
        if (i!=j&&b[j].l<=b[i].l&&b[j].r>=b[i].r) {flag=0;break;}
        if (flag) c[++cnt]=b[i];
    }
    m=cnt;sort(c+1,c+m+1);
    cnt=0;
    for (int i=1;i<=m;i++)
    {
        int t=i;
        while (t<m&&c[t+1].l<=c[t].r) t++;
        cnt++;b[cnt].l=c[i].l,b[cnt].r=c[t].r;
        i=t;
    }
    m=cnt;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
        delta[b[j].l-a[i]+K]++,delta[b[j].r+1-a[i]+K]--;
    cnt=0;
    for (int i=0;i<(K<<1);i++)
    {
        cnt+=delta[i];
        if (cnt>s||cnt==s&&abs(i-K)<d) d=abs(i-K),s=cnt;
    }
    cout<<d<<' '<<s;
    return 0;
}