The second line contains n+mn+m integers x1,x2,…,xn+mx1,x2,…,xn+m (1≤x1<x2<…<xn+m≤1091≤x1<x2<…<xn+m≤109 ), where xixi is the coordinate where the ii -th resident lives.

The third line contains n+mn+m integers t1,t2,…,tn+mt1,t2,…,tn+m (0≤ti≤10≤ti≤1 ). If ti=1ti=1 , then the ii -th resident is a taxi driver, otherwise ti=0ti=0 .

It is guaranteed that the number of ii such that ti=1ti=1 is equal to mm .

Print mm integers a1,a2,…,ama1,a2,…,am , where aiai is the answer for the ii -th taxi driver. The taxi driver has the number ii if among all the taxi drivers he lives in the ii -th smallest coordinate (see examples for better understanding).

3 1
1 2 3 10
0 0 1 0

3 

3 2
2 3 4 5 6
1 0 0 0 1

2 1 

1 4
2 4 6 10 15
1 1 1 1 0

0 0 0 1 


看到這道題感覺很簡單就是寫不出來，隊友都可快做出來了，後來看了大佬程式碼，真是美如畫。

正反遍歷預處理每個點的最近司機，然後再遍歷判斷即可

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
using namespace std;
#define debug(x) cout << "&&" << x << "&&" << endl;
#define lowbit(x) (x&-x)
#define mem(a,b) memset(a, b, sizeof(a));
typedef long long ll;
const ll mod=1000000007;
const int inf = 0x3f3f3f3f;
void read() {freopen("in.txt","r",stdin);}
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//head

const int maxn=200010;
int n,m,L[maxn],R[maxn],ans[maxn],pos[maxn],dir[maxn];
int main() {
    //read();
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n+m;i++) scanf("%d",&pos[i]);
    for(int i=1;i<=n+m;i++) scanf("%d",&dir[i]);
    for(int i=1;i<=n+m;i++) {//從1開始，很巧妙，預處理也很巧
        if(dir[i]) L[i]=i;
        else L[i]=L[i-1];
    }
    R[n+m+1]=n+m+1;
    for(int i=n+m;i>=1;i--) {
        if(dir[i]) R[i]=i;
        else R[i]=R[i+1];
    }
    for(int i=1;i<=n+m;i++) {
        if(!dir[i]) {//如果是乘客，並且L[i]為0即沒有左司機，或右司機>n+m並且左距離大於右距離
            if(!L[i]||(R[i]<=n+m&&pos[i]-pos[L[i]]>pos[R[i]]-pos[i])) ans[R[i]]++;
            else ans[L[i]]++;
        }
    }
    for(int i=1;i<=n+m;i++) {
        if(dir[i]) printf("%d ",ans[i]);
    }
}