Disjoint-set exchange plaid --Codeforces 28B

Description [title] One day an array of n squares decided to play a game. Initially each equal to the number stored in the lattice grid number (from 1
starts). Each grid to determine their own favorite number di, process each round of game play, any of a lattice grid of i and j meet if | i - j | =
di, you can exchange their digital store. Operation sequence and number of times limitation. You will know each grid favorite number, from 1 to n and an
arrangement. You need to determine whether the game can be arranged to the specified state. [Input format input of the first row contains an integer n-(. 1 <= n-<=
100), indicates the number of grid array. (From the initial state 1-n) comprises a second row of n different integers in size from 1 to n, the arrangement specified. (Final state reached)
a third line contains n integers in size from 1 to n, indicates that the corresponding lattice favorite number. [] If the output format to the specified game can be aligned, output YES, otherwise the output
NO.

The title key is converted meaning of the questions:
a start number at position i is i, the last number in position i is j (pos [j] = i )
If the two numbers can be exchanged, or by several other indirect exchange, that the end state can be achieved.

abs(I-j) = num
(1) I > j : I - j = num , j = I - num
(2) I < j : j - I = num, j = I + num

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<vector>
#include<string>
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int maxn = 105;
int n;
int pre[maxn];
int pos[maxn];
int fav[maxn];

void init(){
    for(int i=0; i<=n; i++)
        pre[i] = i;
}
int find(int x){
    int r = x;
    while(r!=pre[r])
        r = pre[r];
    while(x!=r){
        int t = pre[x];
        pre[x] = r;
        x = t;
    }
    return r;
}
void join(int a,int b){
    int fa = find(a);
    int fb = find(b);
    if(fa!=fb)
        pre[fa] = fb;
}
void in_put(){
    int t;
    for(int i=1; i<=n; i++){
        cin >> t;
        pos[t] = i; //t最后要在第i的位置——要看能否与i联通起来
    }
    for(int i=1; i<=n; i++)
        cin >> fav[i];
}
void Union(){
    //位置i 喜欢的数是fav[i]
    for(int i=1; i<=n; i++){
        if(i+fav[i]<=n)
            join(i,fav[i]+i);
        if(i-fav[i]>=1)
            join(i,i-fav[i]);
    }
}
void check(){
    bool flag = 1;
    for(int i=1; i<=n; i++){
        //这个位置的格子不能与最后在这个位置的格子联通
        if(find(i) != find(pos[i])){
            flag = 0;
            break;
        }
    }
    if(flag)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}
int main(){
    while(cin >> n){
        init();
        in_put();
        Union();
        check();
    }
    return 0;
}