codeforce 285 div2 d 题解

摘要: codeforce 285 div2 d 题解

#codeforce 285 div2 D 题解

##说明
这道题目是看了思路分析才知道的，关键问题在于数据量过大，需要快速检索的同时不能辅助空间过大.
因此利用了下面3种方法结合解决该问题

康拓展开与逆康拓展开

树状数组

二分查找

##代码

/**
* @brief Codeforces Round #285 (Div. 2) d
* @file d.cpp
* @author mianma
* @created 2015/01/27 18:18
* @edited  2015/01/29 22:45
* @type math tree
*/
#include <fstream>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <vector>
#include <set>
#include <utility>
#include <stack>

using namespace std;

#define max(a, b)  ((a) > (b) ? (a) : (b))
#define min(a, b)  ((a) > (b) ? (b) : (a))
#define abs(a)     ((a) >  0  ? (a) : (0 - (a)))
#define CLR(vec)   memset(vec, 0, sizeof(vec))
#define PI         acos(-1.0)

#ifdef DEBUG
ifstream in;
ofstream out;
#define CIN in
#define COUT out
#else
#define CIN cin
#define COUT cout
#endif

/*api for binary index tree*/
static inline int low_bit(int i){
return ( i & ( - i));
}

static inline void build_bit(int *vec, int *bit, int size){
for(int i = 1; i <= size; i++){
bit[i] = vec[i];
for(int j = i - 1; j > i - low_bit(i); j-= low_bit(j))
bit[i] += bit[j];
}
}

static inline int sum_bit(int *bit, int k){
int ans = 0;
for(int i = k; i > 0; i -= low_bit(i))
ans += bit[i];
return ans;
}

static inline void update_bit(int *bit, int size, int i, int val){
for(int j = i; j <= size; j += low_bit(j))
bit[j] += val;

}

/* O(lgn) search, O(lgn) update , O(n) memory cost*/

static inline void debug_vec(int *vec, int size){
for(int i = 0; i < size; i++)
COUT << vec[i] << " ";
COUT << "\n";
}

#define MAXN 200010

int n;
int p[MAXN];
int q[MAXN];
int bit[MAXN];

int main(void){
ios_base::sync_with_stdio(0);
#ifdef DEBUG
CIN.open("./in",  ios::in);
COUT.open("./out",  ios::out);
#endif

/* Cantor expansion */
CIN >> n;
int tmp;
CLR(bit);
for(int i = 1; i <= n; i++){
CIN >> tmp;
tmp++;
update_bit(bit, n, tmp, 1);
p[i] = tmp - sum_bit(bit, tmp);
}

#ifdef DEBUG
COUT << "vec for p\n";
debug_vec(p + 1, n);
#endif

CLR(bit);
for(int i = 1; i <= n; i++){
CIN >> tmp;
tmp++;
update_bit(bit, n, tmp, 1);
q[i] = tmp - sum_bit(bit, tmp);
}

#ifdef DEBUG
COUT << "vec for q\n";
debug_vec(q + 1, n);
#endif

for(int i = n; i >= 1; i--){
p[i] += q[i];
p[i - 1] += p[i]/(n - i + 1);
p[i] = p[i]%(n - i + 1);
}

#ifdef DEBUG
COUT << "vec for p + q\n";
debug_vec(p + 1, n);
COUT << endl;
#endif

/*record nums not used*/
CLR(bit);
for(int i = 1; i <= n; i++)
update_bit(bit, n, i, 1);

/* reverse Cantor expansion */
for(int i = 1; i <= n; i++){
int val = p[i] + 1;
int lft = val;
int rht = n;

/*binary search*/
while(lft < rht){
int mid = ( (rht - lft)>> 1) + lft;
int top = sum_bit(bit, mid);
if(top < val){
lft = mid + 1;
}else{
rht = mid;
}
}
q[i] = lft;
update_bit(bit, n, lft, -1);
}

for(int i = 1; i <= n; i++)
COUT << q[i] - 1 << (i == n ? "\n" : " ");
return 0;
}