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  • Gym

    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;
    }
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  • 原文地址:https://www.cnblogs.com/war1111/p/11199791.html
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