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

130 1 2

 

Sample Output

3 1 2

 

题目链接:点击打开链接

可以想到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[n]);}return 0;}