http://acm.hdu.edu.cn/showproblem.php?pid=4911

给定一个序列，有k次机会交换相邻两个位置的数，问说最后序列的逆序对数最少为多少。

实际上每交换一次能且只能减少一个逆序对，所以问题转换成如何求逆序对数。

归并排序或者树状数组都可搞

树状数组：

先按大小排序后分别标号，然后就变成了求1~n的序列的逆序数，每个分别查询出比他小的用i减，在把他的值插入即可

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define clr0(x) memset(x,0,sizeof(x))
typedef long long LL;
typedef pair<int,int> p;const int maxn=100005;
LL f[maxn];
int n;
void add(int x,LL y)
{for(;x<=n;x += x&(-x)) f[x]+=y;
}
LL sum(int x){LL s=0;for (;x;x -= x&(-x)) s+=f[x];return s;
}p a[maxn];
int k;bool cmp(p i,p j){return i.second < j.second;
}int main(){int i;LL s;while (~RD2(n,k)){for (i=0;i<n;++i){RD(a[i].first);a[i].second=i;}sort(a,a+n);for (i=0;i<n;++i)a[i].first = i+1;sort(a,a+n,cmp);clr0(f);for (s=i=0;i<n;++i){s += i - sum(a[i].first);add(a[i].first,1);}printf("%I64d\n",max(0LL,s-k));}return 0;
}

归并排序：

每次归并发现需要前后交换时都给总的ret加上mid - mvl + 1,因为mvl到mid直接的数都比mvr下标上的数大

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
typedef long long LL;
const int maxn = 1e5+5;
LL k;
int n, a[maxn], b[maxn];LL merge_sort(int l, int r)
{if (l == r)return 0;int mid = (l + r) / 2;LL ret = merge_sort(l, mid) + merge_sort(mid+1, r);int mvl = l, mvr = mid+1, mv = l;while (mvl <= mid || mvr <= r) {if (mvr > r || (mvl <= mid && a[mvl] <= a[mvr])) {b[mv++] = a[mvl++];} else {ret += mid - mvl + 1;b[mv++] = a[mvr++];}}for (int i = l; i <= r; i++)a[i] = b[i];return ret;
}int main () {while (scanf("%d%I64d", &n, &k) == 2) {for (int i = 1; i <= n; i++)RD(a[i]);printf("%I64d\n", max(merge_sort(1, n) - k, 0LL));}return 0;
}