BZOJ2621 [Usaco2012 Mar]Cows in a Skyscraper

首先比较容易想到是状态压缩DP

令$f[S]$表示选取了集合$S$以后，已经送了最少次数$cnt$且当前电梯剩下的体积$rest$最大(即$f[S]$是一个二元组$(cnt, rest)$)

于是$f[S] = min_{i \in S} f[S - {i}] + v[i]$

重载的$<$和$+$运算详情就请看程序好了，反正就是一个贪心思想，总复杂度$O(n * 2 ^ {n - 1})$

 /**************************************************************
     Problem: 2621
     User: rausen
     Language: C++
     Result: Accepted
     Time:532 ms
     Memory:13092 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>

 using namespace std;
 const int N = ;
 const int S =  << N;
 const int inf = 1e9;

 int n, mx, mxs;
 int a[S];

 struct data {
     int cnt, rest;
     data(int _c = , int _r = mx) : cnt(_c), rest(_r) {}

     inline data operator + (int t) const {
         static data res;
         res = *this;
         res.rest -= t;
         if (res.rest < ) ++res.cnt, res.rest += mx;
         return res;
     }

     inline bool operator < (const data &d) const {
         return cnt == d.cnt ? rest < d.rest : cnt < d.cnt;
     }
 } f[S];

 int main() {
     int i, s, t, now;
     scanf("%d%d", &n, &mx);
     mxs = ( << n) - ;
     for (i = ; i <= n; ++i) scanf("%d", &a[ << i - ]);
     f[] = data(, mx);
     for (s = ; s <= mxs; ++s) {
         t = s, f[s] = data(inf, mx);
         while (t) {
             now = t & (-t);
             f[s] = min(f[s], f[s ^ now] + a[now]);
             t -= now;
         }
     }
     printf("%d\n", f[mxs].cnt + (f[mxs].rest != mx));
     return ;
 }