ZYB's Premutation(有逆序数输出原序列，线段树)

ZYB's Premutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 758    Accepted Submission(s): 359

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：

已知【1,i】的逆序数，让你还原这个区间；
思路：用数组记录第i个数前面比a[i]大的数的个数；
用线段树记录当前区间的空缺数；类似与买车票插队那道；然后从最后一个数开始，往线段树中插；a[i]越大代表这个数越小；就往左树插，插到相应位置这个位置被用过了；也就是0；
为什么要从后往前插；因为如果从前往后假设一组数1,2,3,4;插得结果会是4，3，2，1；那样就错了；倒插是1，2，3，4；
代码:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define T_T while(T--)
#define mem(x,y) memset(x,y,sizeof(x))
#define ll root<<1
#define rr root<<1|1
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define V(x) tree[x]
const int MAXN=50010;
int tree[MAXN<<2],a[MAXN],ans[MAXN];
int nowans;
void pushup(int root){
	V(root)=V(ll)+V(rr);
}
void made(int root,int l,int r){
	V(root)=r-l+1;
	int mid=(l+r)>>1;
	if(l==r)return;
	made(lson);
	made(rson);
}
void query(int root,int l,int r,int v){
	if(l==r){
		nowans=l;
		V(root)=0;
		return;
	}
	int mid=(l+r)>>1;
	if(v>=V(rr))query(lson,v-V(rr));
	else query(rson,v);
	pushup(root);
}
int main(){
	int T,N;
	SI(T);
	T_T{
		SI(N);
		made(1,1,N);
		int cur,last=0;
		for(int i=0;i<N;i++){
			scanf("%d",&cur);
			a[i]=cur-last;
			last=cur;
		}
		for(int i=N-1;i>=0;i--){
			query(1,1,N,a[i]);
			ans[i]=nowans;
		}
		for(int i=0;i<N;i++){
			if(i)P_;
			printf("%d",ans[i]);
		}puts("");
	}
	return 0;
}