Gym - 102082G
https://vjudge.net/problem/2198225/origin
对于数列中任意一个数,要么从最左边到它不递减,要么从最右边到到它不递减,为了满足这个条件,就要移动,而移动的最少步数就是逆序对数。所以这个数要么往左移动,要么往右移动,所以两个取最小就好了
#include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cmath> #include<ctime> #include<set> #include<map> #include<stack> #include<cstring> #define inf 2147483647 #define ls rt<<1 #define rs rt<<1|1 #define lson ls,nl,mid,l,r #define rson rs,mid+1,nr,l,r #define N 1000010 #define For(i,a,b) for(long long i=a;i<=b;i++) #define p(a) putchar(a) #define g() getchar() using namespace std; long long n,len,ans,k; long long a[N],b[N],c[N],temp,d1[N],d2[N]; long long t[N]; void in(long long &x){ long long y=1; char c=g();x=0; while(c<'0'||c>'9'){ if(c=='-')y=-1; c=g(); } while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=g(); } x*=y; } void o(long long x){ if(x<0){ p('-'); x=-x; } if(x>9)o(x/10); p(x%10+'0'); } void modify(long long k){ for(;k<=n;k+=k&(-k)) t[k]++; } long long getnum(long long k){ long long cnt=0; for(;k>0;k-=k&(-k)) cnt+=t[k]; return cnt; } int main(){ in(n); For(i,1,n){ in(a[i]); b[i]=a[i]; c[i]=a[i]; } sort(b+1,b+n+1); sort(c+1,c+n+1); len=unique(b+1,b+n+1)-b-1; For(i,1,n) a[i]=lower_bound(b+1,b+n+1,a[i])-b; For(i,1,n){ modify(a[i]); d1[i]=i-getnum(a[i]); } memset(t,0,sizeof(t)); for(long long i=n;i;i--){ modify(a[i]); d2[i]=n-i-getnum(a[i])+1; } For(i,1,n) ans+=min(d1[i],d2[i]); o(ans); return 0; }