数据范围：

1 ≤ M ≤ N ≤ 2 × $10^5$

代码1：70pts

#define LOCAL
#include<iostream>
#include<cstdio>

using namespace std;
namespace ly
{
    namespace IO
    {
        #ifndef LOCAL
            #define SIZE (1<<20)
            char in[SIZE],out[SIZE],*p1=in,*p2=in,*p3=out;
            #define getchar() (p1==p2&&(p2=(p1=in)+fread(in,1,SIZE,stdin),p1==p2)?EOF:*p1++)
            #define flush() (fwrite(p3=out,1,SIZE,stdout))
            #define putchar(ch) (p3==out+SIZE&&flush(),*p3++=(ch))
            class Flush{public:~Flush(){flush();}}_;
        #endif
        template<typename type>
        inline void read(type &x)
        {
            x=0;bool flag(0);char ch=getchar();
            while(!isdigit(ch)) flag^=ch=='-',ch=getchar();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
            flag?x=-x:0;
        }
        template<typename type>
        inline void write(type x,bool flag=1)
        {
            x<0?x=-x,putchar('-'):0;static short Stack[50],top(0);
            do Stack[++top]=x%10,x/=10;while(x);
            while(top) putchar(Stack[top--]|48);
            flag?putchar('\n'):putchar(' ');
        }
        #ifndef LOCAL
            #undef SIZE
            #undef getchar
            #undef putchar
            #undef flush
        #endif
    }
}using namespace ly::IO;


#include<bits/stdc++.h>

#define int long long
#define LL long long
int m,n;
const int N=2*1e5+10;
int p[N],ans;
bool pd[N];
//char *p1, *p2, buf[N];
//#define nc() (p1 == p2 && (p2 = (p1 = buf) +\
//fread(buf, 1, N, stdin), p1 == p2) ? EOF : *p1 ++ )
//LL read()
//{
//    LL x = 0, f = 1;
//    char ch = nc();
//    while (ch < 48 || ch > 57)
//    {
//        if (ch == '-') f = -1;
//        ch = nc();
//    }
//    while (ch >= 48 && ch <= 57)
//        x = (x << 3) + (x << 1) + (ch ^ 48), ch = nc();
//    return x * f;
//}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL),cout.tie(NULL);
	read(n),read(m);
	for(int i=1;i<=n;i++)
	{
		read(p[i]);
	}
	sort(p+1,p+1+n);
	for(int i=1;i<=m;i++)
	{
		int b;
		read(b);
		int k=lower_bound(p+1,p+1+n,b)-p;
		while(pd[k])	k++;
		if(p[k]<b||k>n)
		{
			write(-1);
			return 0;
		}
		else
		{
			ans+=p[k];
			pd[k]=1;
		}
	}
	write(ans);
	return 0;
}

代码2:100pts：

#define LOCAL
#include<bits/stdc++.h>
#define int long long

using namespace std;
namespace ly
{
	namespace IO
	{
#ifndef LOCAL
#define SIZE (1<<20)
		char in[SIZE],out[SIZE],*p1=in,*p2=in,*p3=out;
#define getchar() (p1==p2&&(p2=(p1=in)+fread(in,1,SIZE,stdin),p1==p2)?EOF:*p1++)
#define flush() (fwrite(p3=out,1,SIZE,stdout))
#define putchar(ch) (p3==out+SIZE&&flush(),*p3++=(ch))
		class Flush{public:~Flush(){flush();}}_;
#endif
		template<typename type>
		inline void read(type &x)
		{
			x=0;bool flag(0);char ch=getchar();
			while(!isdigit(ch)) flag^=ch=='-',ch=getchar();
			while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
			flag?x=-x:0;
		}
		template<typename type>
		inline void write(type x,bool flag=1)
		{
			x<0?x=-x,putchar('-'):0;static short Stack[50],top(0);
			do Stack[++top]=x%10,x/=10;while(x);
			while(top) putchar(Stack[top--]|48);
			flag?putchar('\n'):putchar(' ');
		}
#ifndef LOCAL
#undef SIZE
#undef getchar
#undef putchar
#undef flush
#endif
	}
}using namespace ly::IO;
int m,n;
const int N=2e6+10;
int a[N],b[N];
bool pd[N];
long long ans;
signed main()
{
	read(n),read(m);
	for(int i=1;i<=n;i++)
		read(a[i]);
	for(int i=1;i<=m;i++)
		read(b[i]);
	sort(a+1,a+1+n);
	sort(b+1,b+1+m);
	int k=1;
	for(int i=1;i<=m;i++)
	{
		while(a[k]<b[i]&&k<=n)	k++;
		while(pd[k])k++;
		if(k>n)
		{
			write(-1);
			return 0;
		}
		ans+=a[k];
		pd[k]=1;
	}
	write(ans);
	return 0;
}