https://vjudge.net/problem/2198221/origin

https://vjudge.net/problem/2198221/origin
逆向思维，原题是人出来，我们处理成人进去，算出来每个人的曼哈顿距离，然后从大到小排序，距离长的先入。走的距离+这个人从队伍中走到入口的距离的最小值就是答案

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define ls rt<<1
#define rs rt<<1|1
#define lson ls,nl,mid,l,r
#define rson rs,mid+1,nr,l,r
#define N 5000010
#define For(i,a,b) for(int i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()

using namespace std;
int n,m,r,mid,k,ans;

struct node{
    int x,y,d;
    bool operator < (const node &a)const{
        return d>a.d;
    }
}a[N];

void in(int &x){
    int y=;
    char c=g();x=;
    while(c<''||c>''){
        if(c=='-')y=-;
        c=g();
    }
    while(c<=''&&c>=''){
        x=(x<<)+(x<<)+c-'';c=g();
    }
    x*=y;
}
void o(int x){
    if(x<){
        p('-');
        x=-x;
    }
    if(x>)o(x/);
    p(x%+'');
}
int main(){
    in(n);in(m);in(r);
    For(i,,r){
        in(a[i].x);in(a[i].y);
        if(a[i].y>m)
            a[i].d=a[i].y-m+n-a[i].x+;
        else
            a[i].d=m-a[i].y+n-a[i].x+;
    }
    sort(a+,a+r+);
    For(i,,r)
        ans=max(ans,(k++)+a[i].d);

    o(ans);
    return ;
}