HDU 5592 ZYB's Premutation (树状数组 + 二分 )

ZYB's Premutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1141    Accepted Submission(s): 601

 

Problem Description

ZYB has a premutation P,but he only remeber the reverse log of each prefix of the premutation,now he ask you to
restore the premutation.

Pair (i,j)(i<j) is considered as a reverse log if Ai>Aj is matched.

 

Input

In the first line there is the number of testcases T.

For each teatcase:

In the first line there is one number N.

In the next line there are N numbers Ai,describe the number of the reverse logs of each prefix,

The input is correct.

1≤T≤5,1≤N≤50000

 

Output

For each testcase,print the ans.

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Sample Input

1 3 0 1 2

 

Sample Output

3 1 2

 

Source

BestCoder Round #65

 

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#include<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<set>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define mod 1e9+7
#define ll long long
#define maxn 100005
#define MAX 1000000000
#define ms memset
using namespace std;

int n , seq[maxn];
int tree[maxn];

int lowbit(int x) { return x & (-x) ; }

int sum(int x)
{
    int res=0;
    for(;x>0;res+=tree[x],x-=lowbit(x));
    return res;
}

void add(int x,int d) {  for( ; x<=n ; tree[x]+=d , x+=lowbit(x) ) ;  }
/*
题目大意：给定n个数，
和序列前缀的逆序对，要求还原这个序列。

这题要根据逆序对序列来倒推，
s[i]-s[i-1]就是这个位置放的数的特征。
放完后用树状数组维护，找位置用二分在树状数组中查找。
当然，自己手工模拟一下就知道找的应该是下界。

*/

int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&seq[i]);

        for(int i=1;i<=n;i++) add(i,1);
        int ans[maxn];

        for(int i=n;i>=1;i--)
        {
            int pos=seq[i]-seq[i-1];
            pos=i-pos;
            int l=1,r=n,op=-1;
            while(r>=l)
            {
                int mid=(l+r)>>1;
                if(sum(mid)>=pos)
                {
                    op=mid;
                    r=mid-1;
                }
                else l=mid+1;
            }
            ans[i]=op;
            add(op,-1);
        }
        for(int i=1;i<=n;i++)    printf("%d%c",ans[i],i==n?'\n':' ');
    }
    return 0;
}