Sicily 14514. Bread Sorting

题目

思路
题意是说将一组数排列成另外一组数，排序的方式是每次取出3个连续的的数字，然后abc可以cab，也就整体右移，超出位置的到最左边。
问能否排列成目标数组。
额，实在是不会做。在网上找到了好几个人的题解，稍加重写并比较性能好了。
好像和树状数组有关。
如代码区所示。

代码
第一种方法，用时0.9s：
by Per Austrin

//Sample solution for the Bread Sorting problem in NCPC 2012 by Per Austrin
//Rewrite by Weiping Huang
//0.9s#include <stdio.h>
#include <algorithm>
using namespace std;int n;
char text[600005];
int A[100005], B[100005], C[100005];void Readin(int * num) {gets(text);int sum = 0, k = 0;for (int i = 0; text[i] != '\0'; i++) {if (text[i] == ' ') {num[k++] = sum;sum = 0;}else {sum = 10 * sum + (text[i] & 15);}}num[k] = sum;//for (int i = 0; i < n; i++) scanf("%d", num + i);
}int inversions(int n, int * N) {if (n < 2) return 0;int m = n / 2;int r = inversions(m, N) + inversions(n - m, N + m);int x = 0, y = m;while (x < m && y < n) {if (N[y] < N[x]) {r += m - x;y++;}else x++;}sort(N, N + n);return r;
}int main() {while (~scanf("%d\n", &n)) {Readin(A);Readin(B);for (int i = 0; i < n; i++) C[B[i]] = i;for (int i = 0; i < n; i++) A[i] = C[A[i]];printf("%sossible\n", inversions(n, A) & 1 ? "Imp" : "P");}return 0;
}

第二种方法，用时0.06s：
By Lukáš Poláček (lukasP)

//Sample solution for the Bread Sorting problem in NCPC 2012 by Lukáš Poláček (lukasP)
//Rewrite by Weiping Huang
//0.06s#include <stdio.h>
#include <string.h>int n;
char text[600005];
bool seen[100005];
int A[100005], B[100005];void ReadinA(int * num) {gets(text);int sum = 0, k = 0;for (int i = 0; text[i] != '\0'; i++) {if (text[i] == ' ') {num[k++] = sum;sum = 0;}else {sum = 10 * sum + (text[i] & 15);}}num[k] = sum;//for (int i = 0; i < n; i++) scanf("%d", num + i);
}void ReadinB(int * num) {gets(text);int sum = 0, k = 0;for (int i = 0; text[i] != '\0'; i++) {if (text[i] == ' ') {num[A[k++] - 1] = sum;sum = 0;}else {sum = 10 * sum + (text[i] & 15);}}num[A[k] - 1] = sum;//for (int i = 0; i < n; i++) scanf("%d", num + i);
}int main() {while (~scanf("%d\n", &n)) {ReadinA(A);ReadinB(B);memset(seen, false, sizeof(bool) * n);int swaps = n;for (int i = 0; i < n; i++) {if (!seen[i]) {swaps--;for (int j = i; !seen[j]; j = B[j] - 1) seen[j] = true;}}printf("%sossible\n", swaps & 1 ? "Imp" : "P");}return 0;
}

第三种方法，用时0.51s：
By Lukáš Poláček (lukasP)

//Sample solution for the Bread Sorting problem in NCPC 2012 by Lukáš Poláček (lukasP)
//Rewrite by Weiping Huang
//0.51s#include <stdio.h>
#include <string.h>
#include <math.h>int n;
char text[600005];
int A[100005], B[100005];struct Buckets {int n, b_size;int s[100005];bool t[100005];void clear(int n) {b_size = int(sqrt((double)n) + 1.1);for (int i = (n + b_size - 1) / b_size - 1; i >= 0; i--) s[i] = 0;for (int i = n - 1; i >= 0; i--) t[i] = false;}void update(int pos) {t[pos] = true;s[pos / b_size]++;}int query(int val) {int b_ind = val / b_size;int res = 0;for (int i = 0; i < b_ind; i++) res += s[i];for (int i = b_ind * b_size; i < val; i++) res += t[i];return res;}
}s, t;void Readin(int * num) {gets(text);int sum = 0, k = 0;for (int i = 0; text[i] != '\0'; i++) {if (text[i] == ' ') {num[k++] = sum - 1;sum = 0;}else {sum = 10 * sum + (text[i] & 15);}}num[k] = sum - 1;//for (int i = 0; i < n; i++) scanf("%d", num + i);
}int main() {while (~scanf("%d\n", &n)) {long long ia = 0, ib = 0;s.clear(n);t.clear(n);memset(A, 0, sizeof(int) * n);memset(B, 0, sizeof(int) * n);Readin(A);for (int i = 0; i < n; i++) {ia += i - s.query(A[i]);s.update(A[i]);}Readin(B);for (int i = 0; i < n; i++) {ib += i - t.query(B[i]);t.update(B[i]);}printf("%sossible\n", ((ia & 1) != (ib & 1)) ? "Imp" : "P");}return 0;
}

第四种方法，用时0.33s：
By Andreas Björklund

//Sample solution for the Bread Sorting problem in NCPC 2012 by Andreas Björklund
//Rewrite by Weiping Huang
//0.33s#include <stdio.h>
#include <string.h>
#include <math.h>int n, tpn;
char text[600005];
int A[1 << 17], B[1 << 17];
int sum1, sum2;int count(int* T, int who)
{int ix = who;int cnt = 0;while (ix <= tpn) {int inc = ix & -ix;cnt ^= (ix <= who) ? T[ix - 1] : 0;T[ix - 1] ^= (ix >= who);ix ^= inc;ix |= (inc << 1);}return cnt & 1;
}void ReadinAndCount() {sum1 = sum2 = 0;gets(text);int sum = 0;for (int i = 0; text[i] != '\0';) {if (text[i] == ' ') {i++;while (text[i] == ' ') i++;sum1 ^= count(A, sum);sum = 0;}else {sum = 10 * sum + (text[i] & 15);i++;}}sum1 ^= count(A, sum);sum = 0;gets(text);for (int i = 0; text[i] != '\0';) {if (text[i] == ' ') {i++;while (text[i] == ' ') i++;sum2 ^= count(B, sum);sum = 0;}else {sum = 10 * sum + (text[i] & 15);i++;}}sum2 ^= count(B, sum);sum = 0;
}int main() {while (~scanf("%d\n", &n)) {tpn = 1;while (tpn < n) tpn <<= 1;for (int i = 0; i < n; i++) A[i] = B[i] = 0;ReadinAndCount();printf("%sossible\n", sum1 != sum2 ? "Imp" : "P");}return 0;
}

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