HDU 4911 Inversion 解题报告（逆序数）

Inversion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 261    Accepted Submission(s): 112

Problem Description

bobo has a sequence a₁,a₂,…,a_n. He is allowed to swap twoadjacent numbers for no more than k times.

Find the minimum number of inversions after his swaps.

Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and a_i>a_j.

 

Input

The input consists of several tests. For each tests:

The first line contains 2 integers n,k (1≤n≤10⁵,0≤k≤10⁹). The second line contains n integers a₁,a₂,…,a_n (0≤a_i≤10⁹).

 

Output

For each tests:

A single integer denotes the minimum number of inversions.

 

Sample Input

3 12 2 13 02 2 1

 

Sample Output

12

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 

    解题报告：求逆序数inv，输出max(inv-k, 0)即可。这里用线段树求逆序数。代码如下：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <iomanip>using namespace std;#define ff(i, n) for(int i=0;i<(n);i++)#define fff(i, n, m) for(int i=(n);i<=(m);i++)#define dff(i, n, m) for(int i=(n);i>=(m);i--)#define bit(n) (1LL<<(n))typedef long long LL;typedef unsigned long long ULL;const LL inf=1e15;void work();int main(){#ifdef ACM    freopen("input.in", "r", stdin);#endif // ACM    work();}/***************************************************/#define lson l, m, pos<<1#define rson m+1, r, pos<<1|1#define defm int m = (l+r)/2;const int maxn = 111111;int num[111111];int yy[111111];int n, k;LL sum[maxn << 2];void updateFather(int pos){    sum[pos] = sum[pos<<1] + sum[pos<<1|1];}void update(int p, int l, int r, int pos){    if(l == r)    {        sum[pos]++;        return;    }    defm;    if(p <= m)        update(p, lson);    else        update(p, rson);    updateFather(pos);}LL query(int L, int R, int l, int r, int pos){    if(L<=l && r<=R)        return sum[pos];    defm;    return (L<=m?query(L, R, lson):0) + (m<R?query(L, R, rson):0);}void work(){    while(scanf("%d%d", &n, &k) == 2)    {        ff(i, n) scanf("%d", num + i);        memcpy(yy, num, sizeof(num));        sort(yy, yy+n);        int tot = unique(yy, yy+n) - yy;        LL ans = 0;        memset(sum, 0, sizeof(sum));        ff(i, n)        {            int pos = lower_bound(yy, yy+tot, num[i]) - yy;            ans += query(pos+1, tot, 0, tot, 1);            update(pos, 0, tot, 1);        }        cout << max(ans - k, 0LL) << endl;    }}