做题记录整理dp1 P1541. [NOIP2010 提高组] 乌龟棋（2022/9/20）

P1541. [NOIP2010 提高组] 乌龟棋

把每个牌选多少个塞进dp的四个维度里，就可以做到无后效性了

#include<bits/stdc++.h>
using namespace std;
#define for1(i,a,b) for(ll i = a;i <=b;i++)
#define ll long long
ll n,a[50000],p[10],x,m,dp[50][50][50][50];
int main()
{
	int x,y,k,ji;
    scanf("%d%d",&n,&m);
	for1(i,1,n)scanf("%d",&a[i]);
	for1(i,1,m) scanf("%d",&x),p[x]++;
	dp[0][0][0][0]=a[1];
	for1(i1,0,p[1])
	{
		for1(i2,0,p[2])
		{
			for1(i3,0,p[3])
			{
				for1(i4,0,p[4])
				{
				int ji=i1+i2*2+i3*3+i4*4+1;
					if(i1!=0) dp[i1][i2][i3][i4]=max(dp[i1][i2][i3][i4],dp[i1-1][i2][i3][i4]+a[ji]);
					if(i2!=0) dp[i1][i2][i3][i4]=max(dp[i1][i2][i3][i4],dp[i1][i2-1][i3][i4]+a[ji]);
					if(i3!=0) dp[i1][i2][i3][i4]=max(dp[i1][i2][i3][i4],dp[i1][i2][i3-1][i4]+a[ji]);
					if(i4!=0) dp[i1][i2][i3][i4]=max(dp[i1][i2][i3][i4],dp[i1][i2][i3][i4-1]+a[ji]);
					
				}
			}
		}
	}
	printf("%d",dp[p[1]][p[2]][p[3]][p[4]]);
	return 0;
 }