HDU-4911-inversion (discretization + tree array)

Question: a sequence composed of N non-negative integers, which is the smallest reverse logarithm (1 ≤ n ≤ 10 ^ 5) after a maximum of K adjacent exchanges, 0 ≤ k ≤ 10 ^ 9, 0 ≤ AI ≤ 10 ^ 9 )..

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4911

--> Only two adjacent numbers can be exchanged at a time, and only the reverse order of the two numbers is changed at a time. Other numbers do not change the reverse order of the two numbers. Therefore, find all the reverse order pairs, and subtract K from them to be the answer.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 100000 + 10;int n, k;int a[MAXN], b[MAXN];long long C[MAXN];void Init(){    memset(C, 0, sizeof(C));}void Read(){    for (int i = 0; i < n; ++i)    {        scanf("%d", a + i);    }}int Lowbit(int x){    return x & (-x);}long long Sum(int x){    long long nRet = 0;    while (x > 0)    {        nRet += C[x];        x -= Lowbit(x);    }    return nRet;}void Add(int x){    while (x <= n)    {        C[x]++;        x += Lowbit(x);    }}void Solve(){    int nCnt = 0;    int nId = 0;    long long nReverse = 0;    memcpy(b, a, sizeof(a));    sort(b, b + n);    nCnt = unique(b, b + n) - b;    for (int i = n - 1; i >= 0; --i)    {        nId = lower_bound(b, b + nCnt, a[i]) - b + 1;        nReverse += Sum(nId - 1);        Add(nId);    }    if (nReverse > k)    {        nReverse -= k;    }    else    {        nReverse = 0;    }    printf("%I64d\n", nReverse);}int main(){    while (scanf("%d%d", &n, &k) == 2)    {        Init();        Read();        Solve();    }    return 0;}

HDU-4911-inversion (discretization + tree array)