【P1064 金明的预算方案】 Dp

P1064
其实早就想到解法了 没看到每个主件只能有 0 1 2 个附件
就心态崩了
于是你可以先分个组
然后当0 1 背包跑就行了
转移的时候几个方案分别选择
总共有 4 个方案
取这个主件/ 取这个主件和他第一个附件/取这个主件和他第二个附件/取这个主件和他所有附件

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
int dp[32025],f[62][3],p[62][3];

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int V,n,v,w,P,c,ans = 0;scanf("%d%d",&V,&n);
    V/=10;
    for(int i = 1;i<=n;++i)
    {
        scanf("%d%d%d",&P,&w,&v);
        if(v==0) v = i;
        for(int j = 0;j<=2;++j)
        {
            if(f[v][j]==0)
            {
                f[v][j] = P/10;
                p[v][j] = w*P/10;
                break;
            }
        }
    }
    for(int i = 1;i<=n;++i)
    {
        if(!f[i][0]) continue;
        int v0 = f[i][0],v1 = f[i][1],v2 = f[i][2];
        int p0 = p[i][0],p1 = p[i][1],p2 = p[i][2];
        for(int j = V;j>=f[i][0];j--)
        {
            dp[j] = max(dp[j],dp[j-f[i][0]] + p[i][0]);
            if(j>=v0+v1&&dp[j-v0-v1]+p0+p1>dp[j]) dp[j] = dp[j-v0-v1]+p0+p1;
            if(j>=v0+v2&&dp[j-v0-v2]+p0+p2>dp[j]) dp[j] = dp[j-v0-v2]+p0+p2;
            if(j>=v0+v2+v1&&dp[j-v0-v2-v1]+p0+p2+p1>dp[j]) dp[j] = dp[j-v0-v2-v1]+p0+p1+p2;
        }
    }
    for(int i = 1;i<=V;++i)
        ans = max(ans,dp[i]);
    printf("%d\n",ans*10);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}