GYM101002C - Greetings!

做法：\(dp[i][S]\)表示用了\(i\)种信封，明信片的状态为\(S\)时的最小花费，预处理\(A[S]\)表示可以装的对应明信片的集合的花费, \(dp[i][s] =min(dp[i-1][s-s2] + A[s2]), s2 \in s\)。注意枚举所有子集，再对每个子集枚举他们的子集，复杂度是\(O(3^n)\)，证明就是 \(O(\sum_{i=0}^n C_n^i 2^i) = O(\sum_{i=0}^n C_n^i 2^i 1^{n-i}) = O((2+1)^n) = O(3^n)\)
...学到了

#include <bits/stdc++.h>
#define pb push_back
typedef long long ll;
const int N = 20;
const ll inf = 0x3f3f3f3f3f3f3f;
using namespace std;
int n,k;
ll w[N], h[N], q[N];
struct node{ll w,h,c;
}A[(1<<16)];
ll dp[N][(1<<16)];
ll min(ll a,ll b) {if(a==-1) return b;if(b==-1) return a;if(a<b) return a;return b;
}
int main() {scanf("%d%d",&n,&k);for(int i = 0; i < n ; ++i) scanf("%lld%lld%lld",&w[i],&h[i],&q[i]);for(int st = 0; st < (1<<n); ++st) {ll tmp = 0, num = 0;A[st].w = A[st].h = 0;for(int i = 0; i < n; ++i) if(st&(1<<i)) {A[st].w = max(A[st].w, w[i]);A[st].h = max(A[st].h, h[i]);tmp += w[i]*h[i]*q[i];num += q[i];}A[st].c = A[st].w*A[st].h*num - tmp;}memset(dp,-1,sizeof(dp));dp[0][0] = 0;for(int i = 1; i <= k; ++i) {for(int st = 0; st < (1<<n); ++st) {ll tmp = -1;for(int st2 = st; st2; st2=(st2-1)&st) if(dp[i-1][st-st2]!=-1) {tmp = min(dp[i-1][st-st2] + A[st2].c,tmp);}dp[i][st] = tmp;}}ll ans = -1;for(int i = 1; i <= k; ++i) ans = min(ans, dp[i][(1<<n)-1]);printf("%lld\n",ans);return 0;
}