hdoj 5592 ZYB's Premutation 【线段树插空】

ZYB's Premutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 31    Accepted Submission(s): 8

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

130 1 2

 

Sample Output

3 1 2

 

题意：给定1-n某一个全排列的前缀区间逆序对个数，让你还原这个序列。

思路：线段树插空，逆向跑一遍就ok了。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (50000+10)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;int a[MAXN], b[MAXN];struct Tree{    int l, r, len;    int sum;};Tree tree[MAXN<<2];void PushUp(int o){    tree[o].sum = tree[ll].sum + tree[rr].sum;}void Build(int o, int l, int r){    tree[o].l = l; tree[o].r = r;    tree[o].len = r - l + 1;    tree[o].sum = 1;    if(l == r)        return ;    int mid = (l + r) >> 1;    Build(lson); Build(rson);    PushUp(o);}int Query(int o, int v){    if(tree[o].l == tree[o].r)        return tree[o].l;    int mid = (tree[o].l + tree[o].r) >> 1;    if(tree[ll].sum >= v)        return Query(ll, v);    else        return Query(rr, v-tree[ll].sum);}void Update(int o, int P){    if(tree[o].l == tree[o].r)    {        tree[o].sum = 0;        return ;    }    int mid = (tree[o].l + tree[o].r) >> 1;    if(P <= mid)        Update(ll, P);    else        Update(rr, P);    PushUp(o);}int main(){    int t; Ri(t);    W(t)    {        int n; Ri(n);        for(int i = 1; i <= n; i++)            Ri(a[i]);        a[0] = 0; Build(1, 1, n);        for(int i = n; i >= 1; i--)        {            int num = a[i] - a[i-1];            int P = Query(1, num+1);            //Pi(P);            Update(1, P);            b[i] = n-P+1;        }        for(int i = 1; i <= n; i++)        {            if(i > 1) printf(" ");            printf("%d", b[i]);        }        printf("\n");    }    return 0;}