$f[s]$表示$s$状态下的最小代价,$g[s]$表示$s$状态下的最大剩余体积

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while(isdigit(ch)) {x = 10*x+ch-'0'; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) putchar('-'),x = -x;
    if(x > 9) out(x/10);
    putchar(x%10+'0');
}
//-------------------------------------------------------------

int n,w,a[20],f[1<<20],g[1<<20];

int main() {
    int i,j; in(n); in(w);
    for(i = 0;i < n; ++i) in(a[i]);
    memset(f,0x3f,sizeof(f));
    f[0] = 0;
    g[0] = 0;
    for(i = 1;i < (1<<n); ++i) {
        for(j = 0;j < n; ++j) {
            if(!(i&(1<<j))) continue;
            if(g[i&~(1<<j)]-a[j] >= 0 && f[i] >= f[i&~(1<<j)]) {
                f[i] = f[i&~(1<<j)];
                g[i] = max(g[i],g[i&~(1<<j)]-a[j]);
            }
            if(g[i&~(1<<j)]-a[j] < 0 && f[i] >= f[i&~(1<<j)]+1) {
                f[i] = f[i&~(1<<j)]+1;
                g[i] = max(g[i],w-a[j]);
            }
        }
    }
    out(f[(1<<n)-1]);
    return 0;
}