rt，这份代码是 $m^2$ 的，题目中说 $m \le 5 \times 10^5$，应该无法通过才对，但它却AC了

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+5;
typedef long long ll;
int n,q;
ll a[N],b[N],c[N],v[N],FC[N],FB[N];
ll m,p[N],DP0[N],mindp=0,mindpid=0;
int main()
{
	//freopen("reserve7.in","r",stdin);
	//freopen("reserve7.out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%lld",&a[i]);
	for(int i=1;i<=n;i++)
	scanf("%lld",&b[i]);
	for(int i=1;i<=n;i++)
	scanf("%lld",&c[i]);
	scanf("%d",&q);
	memset(DP0,0x3f,sizeof(DP0));

	for(int i=1;i<=n;i++)
	{
		FC[i]=FC[i-1]+c[i];
		FB[i]=FB[i-1]+b[i];
	}
	if(FC[n]==0)
	{
		DP0[0]=0;
	
		for(int i=1;i<=n;i++)
		{
			DP0[i]=min(a[i],DP0[i-1]+b[i]);
		}	
	}
	else
	{
		DP0[0]=0;	
		for(int i=1;i<=n;i++)
		{
			DP0[i]=min(a[i],DP0[i-1]+b[i]);
		}		
	}
	
	while(q--)
	{
	
		//memset(ans,0,sizeof(ans));
		scanf("%lld",&m);
		ll ans1=LONG_MAX,ans2=0;
		for(int i=1;i<=m;i++)
		{
			scanf("%lld",&p[i]);	
		}
		if(m==0)
		{
			printf("%lld\n",DP0[n]);
			continue;
		}
		else if(m==1)
		{
			ans1=DP0[n]+c[p[m]];
			ans2=DP0[p[m]-1]+FB[n]-FB[p[m]];
			
			ans1=min(ans1,ans2);
			printf("%lld\n",ans1);
			continue;
		}
		
		else if(FC[n]==0)
		{
			/*int max0=n,i=m;
			while(p[m]==max0) {max0--;m--;}*/
			ans1=DP0[n]+c[p[m]];
			for(int i=1;i<=m;i++)
			{
				ans2=DP0[p[i]-1]+FB[n]-FB[p[i]];
				ans1=min(ans1,ans2);
			}

			printf("%lld\n",ans1);
			continue;	
		}
	
		for(int i=1;i<=m;i++)
		{
			//将1~p[i]-1位置 设为0
			int le=1,ri=p[i]-1;
			ans2=DP0[ri];
			//遍历左边为0的点，单独设1,第i点不用设置
			for(int j=1;j<i;j++)
			{
				ans2+=c[p[j]];
			}
			//遍历右边为1的点，单独设0
			for(int j=i;j<m;j++)
			{
				ans2+=FB[p[j+1]-1]-FB[p[j]];
			}
			//最后到n的点，单独设0
			ans2+=FB[n]-FB[p[m]];
		
			ans1=min(ans1,ans2);
		}
		///全部设0，再单独设置1
		ans2=DP0[n];
		for(int i=1;i<=m;i++)
		ans2+=c[p[i]];
		ans1=min(ans1,ans2);
		printf("%lld\n",ans1);
		
	}
	return 0;
	
	
}