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 = 20;
const int S = 1 << N;
const int inf = 1e9;
 
int n, mx, mxs;
int a[S];
 
struct data {
    int cnt, rest;
    data(int _c = 0, int _r = mx) : cnt(_c), rest(_r) {}
     
    inline data operator + (int t) const {
        static data res;
        res = *this;
        res.rest -= t;
        if (res.rest < 0) ++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 = (1 << n) - 1;
    for (i = 1; i <= n; ++i) scanf("%d", &a[1 << i - 1]);
    f[0] = data(0, mx);
    for (s = 1; 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 0;
}