正题

CF751E

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题目大意

给你一个数组a和一个集合b，现在让你把b中的数插入a，使得逆序对最少

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解题思路

先计算a中的逆序对

对于b和a的逆序对，可以对数字进行排序，用线段树存下放每个位置的最小代价，然后直接求最小值

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code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 1000100
#define mp make_pair
#define fs first
#define sn second
using namespace std;
ll t,n,m,w,x,nm,ans,c[N];
pair<ll,pair<ll,ll> >a[N<<2];
struct Tree
{#define ls x*2#define rs x*2+1ll s[N<<2],lazy[N<<2];void push_up(ll x){s[x]=min(s[ls],s[rs]);return;}void get(ll x,ll y){s[x]+=y;lazy[x]+=y;return;}void push_down(ll x){if(lazy[x]){get(ls,lazy[x]);get(rs,lazy[x]);lazy[x]=0;}return;}void build(ll x,ll l,ll r){s[x]=0;lazy[x]=0;if(l==r)return;ll mid=l+r>>1;build(ls,l,mid);build(rs,mid+1,r);return;}void add(ll x,ll L,ll R,ll l,ll r,ll y){if(L==l&&R==r){get(x,y);return;}push_down(x);ll mid=L+R>>1;if(r<=mid)add(ls,L,mid,l,r,y);else if(l>mid)add(rs,mid+1,R,l,r,y);else add(ls,L,mid,l,mid,y),add(rs,mid+1,R,mid+1,r,y);push_up(x);}ll ask(ll x,ll l,ll r,ll y){if(l==r)return s[x];push_down(x);ll mid=l+r>>1;if(y<=mid)return ask(ls,l,mid,y);else return ask(rs,mid+1,r,y);}
}T;
void add(ll x)
{for(;x<=n;x+=x&-x)c[x]++;return;
}
ll ask(ll x)
{ll sum=0;for(;x;x-=x&-x)sum+=c[x];return sum;
}
int main()
{scanf("%lld",&t);while(t--){scanf("%lld%lld",&n,&m);w=0;nm=n+1;ans=0;T.build(1,1,nm);for(ll i=1;i<=n;++i){c[i]=0;scanf("%lld",&x);T.add(1,1,nm,i+1,nm,1);a[++w]=mp(x,mp(0,i));a[++w]=mp(x,mp(1,i));//相同的数字不计逆序对}for(ll i=1;i<=m;++i){scanf("%lld",&x);a[++w]=mp(x,mp(1,0));}sort(a+1,a+1+w);for(ll i=1;i<=w;++i){if(!a[i].sn.fs){T.add(1,1,nm,a[i].sn.sn+1,nm,-1);ans+=ask(n)-ask(a[i].sn.sn);//计算a中的贡献add(a[i].sn.sn);}else if(!a[i].sn.sn){ans+=T.s[1];}else{T.add(1,1,nm,1,a[i].sn.sn,1);}}printf("%lld\n",ans);}return 0;
}