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;
}

hdu 4911 求逆序对数+树状数组