Taxi drivers and Lyft
Palo Alto is an unusual city because it is an endless coordinate line. It is also known for the office of Lyft Level 5.
Lyft has become so popular so that it is now used by all mm taxi drivers in the city, who every day transport the rest of the city residents — nn riders.
Each resident (including taxi drivers) of Palo-Alto lives in its unique location (there is no such pair of residents that their coordinates are the same).
The Lyft system is very clever: when a rider calls a taxi, his call does not go to all taxi drivers, but only to the one that is the closest to that person. If there are multiple ones with the same distance, then to taxi driver with a smaller coordinate is selected.
But one morning the taxi drivers wondered: how many riders are there that would call the given taxi driver if they were the first to order a taxi on that day? In other words, you need to find for each taxi driver ii the number aiai — the number of riders that would call the ii-th taxi driver when all drivers and riders are at their home?
The taxi driver can neither transport himself nor other taxi drivers.
Input
The first line contains two integers nn and mm (1≤n,m≤1051≤n,m≤105) — number of riders and taxi drivers.
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.
Output
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).
Examples
Input
3 1 1 2 3 10 0 0 1 0
Output
3
Input
3 2 2 3 4 5 6 1 0 0 0 1
Output
2 1
Input
1 4 2 4 6 10 15 1 1 1 1 0
Output
0 0 0 1
Note
In the first example, we have only one taxi driver, which means an order from any of nn riders will go to him.
In the second example, the first taxi driver lives at the point with the coordinate 22, and the second one lives at the point with the coordinate 66. Obviously, the nearest taxi driver to the rider who lives on the 33 coordinate is the first one, and to the rider who lives on the coordinate 55 is the second one. The rider who lives on the 44 coordinate has the same distance to the first and the second taxi drivers, but since the first taxi driver has a smaller coordinate, the call from this rider will go to the first taxi driver.
In the third example, we have one rider and the taxi driver nearest to him is the fourth one.
题目大意：第一行输入两个整数n,m，代表有n个人，m辆出租车，第二行输入n+m个数字，分别是人和出租车的位置，第三行有n+m个数字，1代表是出租车，0代表是人，人会选取离他最近的出租车上车，若其离两辆出租车一样近，那么会选取前一辆出租车，问最终每辆出租车能载几人。
解决方法：此题只需遍历一下当前的人离他前一辆以及后一辆车的距离，及时更新即可。
AC代码：
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
pair<int ,int> p[maxn*3];
int a[maxn];
int num[maxn];
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
ios::sync_with_stdio(0),cin.tie(0);
int n,m;
cin>>n>>m;
rep(i,1,n+m) {
cin>>p[i].first;
}
int an=0;
rep(i,1,n+m) {
cin>>p[i].second;
if(p[i].second==1)
{
a[++an]=p[i].first;
}
}
int k=1;
int i;
for(i=1;i<=n+m;i++) {
if(p[i].second==1)
break;
num[k]++;
}
i++;
for(;i<=n+m;i++)
{
if(p[i].second==1)
{
k++;
continue;
}
if(k<an) {
if(abs(p[i].first-a[k])<=abs(p[i].first-a[k+1]))
num[k]++;
else
num[k+1]++;
}
else
num[k]++;
}
rep(i,1,an) {
cout<<num[i]<<" ";
}
cout<<endl;
return 0;
}