显然相当于使序列变成单峰。给原序列每个数按位置标号，则要求重排后的序列原标号的逆序对数最少。考虑将数从大到小放进新序列，那么贪心的考虑放在左边还是右边即可，因为更小的数一定会在其两侧，与它自身放在哪无关。对于相同的数，一定可以将其安排至无逆序对，特判一下。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; int read() { int x=0,f=1;char c=getchar(); while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();} while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N 300010 int n,tree[N]; long long ans=0; struct data{int x,i; }a[N]; bool cmp(const data&a,const data&b) { return a.x>b.x; } void add(int k){while (k<=n) tree[k]++,k+=k&-k;} int query(int k){int s=0;while (k) s+=tree[k],k-=k&-k;return s;} int main() { #ifndef ONLINE_JUDGE freopen("bzoj4240.in","r",stdin); freopen("bzoj4240.out","w",stdout); const char LL[]="%I64d\n"; #else const char LL[]="%lld\n"; #endif n=read(); for (int i=1;i<=n;i++) a[i].x=read(),a[i].i=i; sort(a+1,a+n+1,cmp); for (int i=1;i<=n;i++) { int t=i;while (t<n&&a[t+1].x==a[i].x) t++; for (int j=i;j<=t;j++) ans+=min(query(a[j].i),i-1-query(a[j].i)); for (int j=i;j<=t;j++) add(a[j].i); i=t; } cout<<ans; return 0; }
原文：https://www.cnblogs.com/Gloid/p/9833852.html