BestCoder Round #65 HDOJ5592 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): 638 Accepted Submission(s): 302



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

题目链接:点击打开链接

能够想到a[i] - a[i - 1]为i前比a[i]大的数的个数, 从后向前遍历, 通过二分与树状数组确定i应当填入的位置.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 5e4 + 5;
int n, a[MAXN], c[MAXN], ans[MAXN];
int lowbit(int x)
{
return x & (-x);
}
void update(int x, int y)
{
while(x <= n) {
c[x] += y;
x += lowbit(x);
}
}
int get_sum(int x)
{
int sum = 0;
while(x > 0) {
sum += c[x];
x -= lowbit(x);
}
return sum;
}
int binary_search(int x)
{
int l = 1, r = n, m;
while(l <= r) {
m = (l + r) >> 1;
if(get_sum(m) >= x) r = m - 1;
else l = m + 1;
}
return l;
}
int main(int argc, char const *argv[])
{
int t;
scanf("%d", &t);
while(t--) {
memset(c, 0, sizeof(c));
memset(ans, 0, sizeof(ans));
scanf("%d", &n);
for(int i = 1; i <= n; ++i)
scanf("%d", &a[i]);
for(int i = n; i >= 1; --i)
a[i] -= a[i - 1];
for(int i = 1; i <= n; ++i)
update(i, 1);
for(int i = n; i >= 1; --i) {
int pos = i - a[i] - 1;
ans[i] = binary_search(pos + 1);
update(ans[i], -1);
}
for(int i = 1; i < n; ++i)
printf("%d ", ans[i]);
printf("%d\n", ans
);
}
return 0;
}