HDU 1394 -- Minimum inversion number (minimum number of reverse orders)

Question:

Given a sequence, the numbers in it are 0 to n-1. Each time we put the first number to the last number and repeat n times, what is the minimum number of reverse orders in N operations?

Ideas:

First, findInitial Number of reverse ordersAnd then move the first number to the last side, then the number of reverse orders minus the number smaller than it, plus the number larger than it. If the number I input is a [I], then the number smaller than it is a [I, the number of numbers larger than it has a n-1-a [I]

Method:Merge Sorting,Tree Array,Line Segment tree

Merge Sort code:

# Include <iostream> # include <cstring> # include <cstdio> # include <algorithm> # define INF 0x3f3f3f3f # define M 5001 using namespace STD; int ans, N; int A [m], t [m], B [m]; void merge_sort (int * a, int X, int y, int * t) {If (Y-x> 1) {int M = (Y + x)> 1; int P = x, q = m, I = x; merge_sort (, x, M, T); merge_sort (a, m, Y, T); While (P <M | q <Y) {If (q> = Y | (P <M & A [p] <= A [Q]) T [I ++] = A [p ++]; else {T [I ++] = A [q ++]; ans + = m-P ;}} for (I = x; I <Y; ++ I) {A [I] = T [I] ;}}int main () {// freopen ("input.txt", "r", stdin); While (scanf ("% d ", & N )! = EOF) {ans = 0; For (INT I = 0; I <n; ++ I) {scanf ("% d", & A [I]); B [I] = A [I]; <span style = "white-space: pre"> </span>} merge_sort (B, 0, N, t ); <span style = "white-space: pre"> </span> // use a temporary array to sort the number of reverse orders, because we also need to reference the number of the original array, int Minn = ans; for (INT I = 0; I <n; ++ I) {ans + = n-1-2 * A [I]; Minn = min (Minn, ANS );} cout <Minn <Endl;} return 0 ;}

Tree array code:

# Include <iostream> # include <cstdio> # include <algorithm> # include <cstring> # define INF 0x3f3f3f3f # define M 5001 using namespace STD; int C [m], A [m], n; int lowbit (int x) {return X &-X;} void Update (int x, int v) {While (x <= N) {c [x] + = V; x + = lowbit (x) ;}} int get_sum (int x) {int ans = 0; while (x> 0) {ans + = C [X]; X-= lowbit (x);} return ans;} int main () {// freopen ("input.txt", "r ", stdin); While (scanf ("% d", & N )! = EOF) {memset (C, 0, sizeof C); int ans = 0; For (INT I = 0; I <n; ++ I) {scanf ("% d", & A [I]); ans + = get_sum (N)-get_sum (A [I] + 1 ); <span style = "white-space: pre"> </span> // calculate the number of updates (A [I] +) larger than a [I );} // cout <ans <Endl; int Minn = inf; For (INT I = 0; I <n; ++ I) {ans + = n-1-2 * A [I]; Minn = min (Minn, ANS);} cout <Minn <Endl;} return 0 ;}

Line Segment Tree Code:

# Include <iostream >#include <cstring> # include <cstdio> # include <algorithm> # define M 5001 # define ls (x) x <1 # define RS (X) x <1 | 1 using namespace STD; int N; int A [m]; struct node {int L, R, num;} tre [4 * m]; void push_up (int rt) {tre [RT]. num = tre [ls (RT)]. num + tre [RS (RT)]. num;} void build (int rt, int L, int R) {tre [RT]. L = L; tre [RT]. R = r; tre [RT]. num = 0; If (L = r) return; int M = (L + r)> 1; build (LS (RT), L, M ); build (RS (RT), m + 1, R); push_up (RT);} void Update (int rt, int P, int v) {If (Tre [RT]. L = tre [RT]. r) {tre [RT]. num = V; return;} int M = (Tre [RT]. L + tre [RT]. r)> 1; if (P <= m) Update (LS (RT), P, V); else Update (RS (RT), P, V ); push_up (RT);} int get_sum (int rt, int L, int R) {If (L <= tre [RT]. L & tre [RT]. r <= r) {return tre [RT]. num;} int sum = 0; int M = (Tre [RT]. L + tre [RT]. r)> 1; if (L <= m) sum + = get_sum (LS (RT), L, R); If (M <R) sum + = get_sum (RS (RT), L, R); R Eturn sum;} int main () {// freopen ("input.txt", "r", stdin); While (scanf ("% d", & N )! = EOF) {build (, n-1); int ans = 0; For (INT I = 0; I <n; ++ I) {scanf ("% d ", & A [I]); ans + = get_sum (1, a [I] + 1, n-1 ); // calculate the number of updates (1, a [I], 1);} int Minn = ans; For (INT I = 0; I <n; ++ I) {ans + = n-1-2 * A [I]; Minn = min (Minn, ANS);} cout <Minn <Endl ;} return 0 ;}