洛谷P1774 

题目：登录 - Luogu Spilopelia​​​​​​

为什么最小的交换次数就是逆序对的个数，请看相关证明

1、归并排序

// 归并排序解法 
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 5e5 + 10;
typedef long long ll;
int a[N],b[N];
int n;
ll ans = 0;
void merge(int l,int r)
{if(l>=r) return;int mid = l + r >> 1;merge(l,mid),merge(mid + 1,r);int k = 0,i = l,j = mid + 1;while(i<=mid&&j<=r)if(a[i]<=a[j]) b[k++] = a[i++];else {ans += mid - i + 1;b[k++] = a[j++];}while(i<=mid)b[k++] = a[i++];while(j<=r) b[k++] = a[j++];for(int i = l,j = 0;i<=r;i++,j++) a[i] = b[j];
}
int main()
{scanf("%d",&n);for(int i = 0;i<n;i++) scanf("%d",&a[i]);merge(0,n-1);printf("%lld\n",ans);return 0;
} 

2、树状数组

进行离散化之后赋予排名，排名的数字越小，则说明数字越小，相同值的话先出现的排名数字越小。然后可以正着也可以反着进行求

// 树状数组的解法
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N = 5e5 + 10;
int tr[N],ranks[N];
int n;
struct Node
{int num,id;
}a[N];
bool cmp(Node &a,Node &b)
{if(a.num!=b.num)return a.num < b.num;return a.id < b.id;
}
int lowbit(int x){return x & -x;
} 
void add(int x,int v)
{for(int i = x;i<=n;i+=lowbit(i))tr[i] += v;
}
ll query(int x)
{ll res = 0;for(int i = x;i;i-=lowbit(i))res += tr[i];return res;
}int main()
{scanf("%d",&n);for(int i = 1;i<=n;i++) {scanf("%d",&a[i].num);a[i].id = i;}sort(a + 1,a + 1 + n,cmp);for(int i = 1;i<=n;i++) ranks[a[i].id] = i;ll ans = 0;// 正向求解 for(int i = 1;i<=n;i++){add(ranks[i],1);ans += i - query(ranks[i]);}/*反向求解 for(int i = n;i>=1;i--){add(ranks[i],1);ans += query(ranks[i]-1);}*/printf("%lld",ans);return 0;
}

3、线段树

离散化之后对每一个排名；

比如说当求到ranks[3] 的时候，看看在它之前有没有 4 - ----n的数字出现，询问以后，把update(ranks[3],1)加到线段树里面去，代表贡献了1.

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1e6;
int ranks[N];
int sum[4*N];
struct Node
{int num,id;
}a[N];
typedef long long ll;
bool cmp(Node &a,Node &b)
{if(a.num!=b.num) return a.num < b.num;return a.id < b.id;
}
void pushup(int x)
{sum[x] = sum[x*2] + sum[x*2+1];
}
void build(int l,int r,int x)
{if(l == r) {sum[x] = 0;return;}int mid = l + r >> 1;build(l,mid,x*2);build(mid + 1,r,x*2 + 1);pushup(x); 
}
void update(int pos,int l,int r,int x,int v)
{if(l == r){sum[x] += v;return;}int mid = l + r >> 1;if(pos<=mid) update(pos,l,mid,x*2,v);else update(pos,mid + 1,r,x*2+1,v);pushup(x);
}
long long query(int pl,int pr,int l,int r,int x)
{if(pl<=l&&r<=pr) return sum[x];int mid = l + r>>1;long long ans = 0;if(pl<=mid) ans += query(pl,pr,l,mid,x*2);if(pr>mid) ans += query(pl,pr,mid + 1,r,x*2+1);return ans;
}
int main()
{int n;scanf("%d",&n);build(1,n,1);for(int i = 1;i<=n;i++){scanf("%d",&a[i].num);a[i].id = i;}sort(a+1,a+1+n,cmp);for(int i = 1;i<=n;i++)ranks[a[i].id] = i;ll ans = 0;n++;for(int i = 1;i<=n-1;i++){ans += query(ranks[i]+1,n,1,n,1);update(ranks[i],1,n,1,1);}printf("%lld",ans);return 0;
}