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


[align=left]Problem Description[/align]
bobo has a sequence a1,a2,…,an. He is allowed to swap two
adjacent 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 ai>aj.
 

[align=left]Input[/align]
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,…,an (0≤ai≤109).
 

[align=left]Output[/align]
For each tests:

A single integer denotes the minimum number of inversions.
 

[align=left]Sample Input[/align]

3 1
2 2 1
3 0
2 2 1

 

[align=left]Sample Output[/align]

1
2

 

[align=left]Author[/align]
Xiaoxu Guo (ftiasch)
 

[align=left]Source[/align]
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;
}
}