最小的字典序就是f1,f2,f3......

這題O(n^2)做法會超時，需要nlongn,套一下模板就好了~

#include<bits/stdc++.h>
using namespace std;
const int inf = 1000000007;
const int N = 100005;
int  a[N], g[N], d[N];  //模板詳見小白書P62

int main(){
    int  t,i,n;
    scanf("%d",&t);
    while(t--){
        memset(d,0,sizeof(d));
        scanf("%d",&n);
        for( i=0; i < n; ++i )scanf("%d", &a[i]);

        for(int i=1; i<=n; i++)g[i]=inf;
        for(int i=0; i<n; i++){
          int k=lower_bound(g+1, g+n+1, a[i])-g;//在g[1]到g[n]中找
          d[i]=k;
          g[k]=a[i];
        }

        for(i=0;i<n;i++)printf("%d%c",d[i],i!=n-1?' ':'\n');
    }
    return 0;
}