codeforces B. Ciel and Duel

B. Ciel and Duel

 

Fox Ciel is playing a card game with her friend Jiro.

Jiro has n cards, each one has two attributes:position (Attack or Defense) and strength. Fox Ciel has m cards, each one has these two attributes too. It's known that position of all Ciel's cards is Attack.

Now is Ciel's battle phase, Ciel can do the following operation many times:

1. Choose one of her cards X. This card mustn't be chosen before.
2. If Jiro has no alive cards at that moment, he gets the damage equal to (X's strength). Otherwise, Ciel needs to choose one Jiro's alive cardY, then:
   • If Y's position is Attack, then (X's strength) ≥  (Y's strength) must hold. After this attack, cardY dies, and Jiro gets the damage equal to (X's strength) - (Y's strength).
   • If Y's position is Defense, then (X's strength) >  (Y's strength) must hold. After this attack, cardY dies, but Jiro gets no damage.

Ciel can end her battle phase at any moment (so, she can use not all her cards). Help the Fox to calculate the maximal sum of damage Jiro can get.

Input

The first line contains two integers n andm (1 ≤ n, m ≤ 100) — the number of cards Jiro and Ciel have.

Each of the next n lines contains a stringposition and an integer strength (0 ≤ strength ≤ 8000) — the position and strength of Jiro's current card. Position is the string "ATK" for attack, and the string "DEF" for defense.

Each of the next m lines contains an integerstrength (0 ≤ strength ≤ 8000) — the strength of Ciel's current card.

Output

Output an integer: the maximal damage Jiro can get.

Sample test(s)

Input

2 3
ATK 2000
DEF 1700
2500
2500
2500

Output

3000

Input

3 4
ATK 10
ATK 100
ATK 1000
1
11
101
1001

Output
992

  这道题我用的是贪心的思路，分两种情况，但是分的还是不够彻底，然后就老是错，后来看了别人的代码，还是给两种情况没有彻底分开了写的，总体来说就是思路不清晰的，
思路：贪心，
分两种情况：
一    对于防御的牌不用管，只考虑战胜对方的atk牌的，这时候就是最基本的贪心，用自己的最大牌去袭击对方的最小牌，取得对方袭击的最大。
二   即去除对方的防御牌，然后在战胜对方的atk牌，然后自己手里剩下的牌就是多可以增大对方的伤害数，所以满足的条件最少是n>m，然后用贪心使尽可能小的牌去战胜对方的防御牌，然后判断手中牌是否可以战胜对方手中剩下的atk牌
取两种方式的最大牌.
下面是代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<climits>
#include<map>
using namespace std;
#define rep(i,n) for(int i=0; i<n; i++)
#define repf(i,n,m) for(int i=(n); i<=(m); ++i)
#define repd(i,n,m) for(int i=(n); i>=(m); --i)
#define fab(a) ((a)>0?(a):(0-(a)))
#define ll long long
#define arc(a) ((a)*(a))
#define inf 100000
#define exp 0.000001
#define N 
vector<int>atk;
vector<int>def;
vector<int>ciel;
vector<bool>sign;
int n,m;
int part()
{
	int sum=0;
	int len=atk.size();
//	cout<<len<<" "<<n<<endl;
//	cout<<"@@@"<<endl;
	int t=min(len,n);
	rep(i,t)
	{
//		cout<<i<<endl;getchar();
		sum+=max(0,ciel[n-1-i]-atk[i]);
	}
//	cout<<sum<<endl;
	return sum;
}
int find(int x)
{
	rep(i,n)
		if(sign[i]==true && ciel[i]>=x)
			return i;
	return -1;
}
int all()
{
	int len=def.size();
	if(len>n)
		return 0;
	int sum=0;
	rep(i,len)
	{
		int x=find(def[i]+1);
		if(x==-1)
			return 0;
		sign[x]=false;
	}
	len=atk.size();
	rep(i,len)
	{
		int x=find(atk[i]);
		if(x==-1)
			return 0;
		sum+=ciel[x];
		sign[x]=false;
	}
	rep(i,n)
		if(sign[i]==true)
			sum+=ciel[i];
    rep(i,len)
		sum-=atk[i];
//	cout<<sum<<endl;
	return sum;
}
int main()
{ 
	char s[5];
	int x;
	while(cin>>m>>n)
	{
		atk.clear();
		def.clear();
		ciel.clear();
		sign.clear();
		rep(i,m)
		{
          cin>>s>>x;
	      if(s[0]=='A')
		     atk.push_back(x);
	      else
		     def.push_back(x);	  
		}
		rep(i,n)
		{
			cin>>x;
			ciel.push_back(x);
			sign.push_back(true);
		}
		sort(atk.begin(),atk.end());
		sort(def.begin(),def.end());
		sort(ciel.begin(),ciel.end());
		int r=max(part(),all());
		cout<<r<<endl;
	}
   return 0;
}