洛谷 P2623 物品选取

https://www.luogu.org/problemnew/show/P2623

https://www.luogu.org/blog/test-1/solution-p2623

重点就是甲类物品最多取一个，一定能取到最优解。。。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<vector>
 5 using namespace std;
 6 #define fi first
 7 #define se second
 8 #define mp make_pair
 9 #define pb push_back
10 typedef long long ll;
11 typedef unsigned long long ull;
12 typedef pair<ll,ll> pll;
13 
14 ll ans[2005],an2;
15 //前i个物品，用j的容量的最大价值
16 vector<pll> dd;
17 ll n,m;
18 int main()
19 {
20     ll idx,i,j,k,a,b,c;
21     scanf("%lld%lld",&n,&m);
22     for(i=1;i<=n;i++)
23     {
24         scanf("%lld",&idx);
25         if(idx==1)
26         {
27             scanf("%lld%lld",&a,&b);
28             dd.pb(mp(a,b));
29         }
30         else if(idx==2)
31         {
32             scanf("%lld%lld%lld",&a,&b,&c);
33             if(b==0)
34             {
35                 for(j=0;j<=m;j++)    ans[j]+=a*c;
36             }
37             else
38             {
39                 for(j=m;j>=1;j--)
40                 {
41                     for(k=min(c,j/b);k>=1;k--)
42                     {
43                         ans[j]=max(ans[j],ans[j-k*b]+k*a);
44                     }
45                 }
46             }
47         }
48         else if(idx==3)
49         {
50             scanf("%lld%lld",&a,&b);
51             if(b==0)    exit(-1);
52             else
53             {
54                 for(j=b;j<=m;j++)
55                 {
56                     ans[j]=max(ans[j],ans[j-b]+a);
57                 }
58             }
59         }
60         //for(int i=0;i<=m;i++)    printf("a%lld %lld\n",i,ans[i]);
61     }
62     for(i=1;i<=m;i++)    ans[i]=max(ans[i],ans[i-1]);
63     an2=ans[m];
64     for(i=0;i<dd.size();i++)
65     {
66         for(j=0;j<=m;j++)
67         {
68             an2=max(an2,ans[m-j]+dd[i].fi*j*j-dd[i].se*j);
69         }
70     }
71     printf("%lld",an2);
72     return 0;
73 }