Bestcoder round #65 && hdu 5592 ZYB's Premutation 线段树

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 175 Accepted Submission(s): 74


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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

[align=left]Output[/align]
For each testcase,print the ans.

[align=left]Sample Input[/align]

1

3
0 1 2

[align=left]Sample Output[/align]

3 1 2

[align=left]Source[/align]
BestCoder Round #65
思路：容易想到k = (a[i] - a[i - 1] + 1)就是原序列第i个数的左边比其大的个数，可以从右往左依次计算

用线段树实现：序列中每个存在的位置相当于有一个1，不存在位0，从中找出第k个1所在的位置，找到后删除该数相当于置0

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <iostream>
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1|1
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long ll;

const int N = 50005;
int num[N << 2], a
, ans
;
void up(int rt) { num[rt] = num[rt << 1] + num[rt << 1|1]; }
void build(int l, int r, int rt)
{
if(l == r) {
num[rt] = 1;
return ;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
up(rt);
}
void update(int l, int r, int rt, int p)
{
if(l == p && r == p) {
num[rt]--;
return ;
}
int m = (l + r) >> 1;
if(p <= m) update(lson, p);
else update(rson, p);
up(rt);
}
int pos;
void get(int l, int r, int rt, int x)
{
if(l == r) {
pos = l;
return;
}
int m = (l + r) >> 1;
if(num[rt << 1] < x) get(rson, x - num[rt << 1]);
else get(lson, x);
}
int main()
{
int _; scanf("%d", &_);
while(_ --)
{
int n; scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
build(1, n, 1);
int c = n;
for(int i = n; i >= 2; --i)
{
int d = c - (a[i] - a[i - 1]);
get(1, n, 1, d);
update(1, n, 1, pos);
//   cout << pos << endl;
ans[i] = pos;
c--;
}
get(1, n, 1, 1); ans[1] = pos;
for(int i = 1; i < n; ++i) printf("%d ", ans[i]);
printf("%d\n", ans
);
}
return 0;
}

View Code