CF and query set (or figure) Ice skating

Is to give you n points, these points must be on the same X or Y axis. The minimum number of points can satisfy this condition.

Ah, I didn't expect it. In fact, if X is the same, y can be moved together.

The first question is wa n times because the array is too small...

/*Pro: 0Sol:date:*/#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <set>#include <vector>using namespace std;int n,a,b,x[1111],y[1111],p[1111];int find(int x) {return x == p[x] ? x : p[x] = find(p[x]);}int main(){    scanf("%d",&n);    for(int i = 1; i <= n; i ++)        scanf("%d%d",&x[i],&y[i]),p[i] = i;    int ans=0;    for(int i = 1; i <= n; i ++){        for(int j = 1; j <= n; j ++){            if(x[i] == x[j] || y[i] == y[j]){                int px = find(i);                int py = find(j);                if(px!=py)                p[px] = py,ans++;            }        }    }    printf("%d\n",n-1-ans);return 0;}

#include <cstdio>int n,x[1111],y[1111],vis[1111],ans;void go(int indx){    vis[indx] = true;//    printf("%d\n",indx);    for(int i = 1; i<= n; i ++)        if( (x[i] == x[indx] || y[i] == y[indx] )&& !vis[i])            go(i);}int main(){    scanf("%d",&n);    for(int i = 1; i <= n; i ++){        scanf("%d%d",x + i, y + i);    }    ans = -1;    for(int i = 1; i <= n; i ++){        if(!vis[i]) go(i),ans ++;    }    printf("%d\n",ans);    return 0;}