HDU 5592 ZYB's Premutation(BestCoder Round #65 C)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5592

ZYB's Premutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 483    Accepted Submission(s): 236

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.

 

 

Sample Input

1 3 0 1 2

 

 

Sample Output

3 1 2

 

 

Source

BestCoder Round #65

 

题意：输入t，代表t组测试样例，每组样例一个n，代表有n个数，输入n个数a[i]，表示从1到i有a[i]个逆序数对，求数字1-n的摆放顺序。

 

分析：线段树维护。

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;


int num[50005];
int ans[50005];
int aa[50005];
struct Tree
{
    int x, sum;
}a[200005];

void Build(int rt, int l, int r, int n)
{
    if(l == r)
    {
        a[rt].x = l;
        a[rt].sum = 1;
        return ;
    }
    Build(rt*2, l, (l+r)/2, n);
    Build(rt*2+1, (l+r)/2+1, r, n);
    a[rt].sum = a[rt*2].sum + a[rt*2+1].sum;
}

int Querry(int rt, int l, int r, int num)
{
    int ans;
    if(l == r)
    {
        a[rt].sum--;
        return a[rt].x;
    }
    if(num > a[rt*2].sum)
        ans = Querry(rt*2+1, (l+r)/2+1, r, num - a[rt*2].sum);
    else
        ans = Querry(rt*2, l, (l+r)/2, num);
    
    a[rt].sum--;
    return ans;
} 

int main()
{
    int t, n;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        aa[0] = 0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &aa[i]);
            num[i] = aa[i] - aa[i-1];//num[i]保存i前面有多少个比ans[i]大
        }
        Build(1, 1, n, n);
        for(int i=n; i>=1; i++)
        {
            ans[i] = Querry(1, 1, n, num[i]+1);
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d%c",ans[i],i==n?'\n':' ');
        }
        
    }
    
    return 0;
}