GYM101002C

GYM101002C - Greetings!

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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
",ans);
    return 0;
}