用f[i]记录状态为i的最小值
g[i] 记录状态为i 的剩下电梯的空间的最大值
#include<bits/stdc++.h>
#define N 19
using namespace std;
int f[1<<N],g[1<<N];
int n,a[N],w;
int main(){
scanf("%d%d",&n,&w); memset(f,127,sizeof(f));
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
f[1<<(i-1)] = 1; g[1<<(i-1)] = w-a[i];
}
for(int i=0;i<=(1<<n)-1;i++) for(int j=1;j<=n;j++){
if((1<<(j-1)) & i) continue;
if(a[j]<=g[i]){
f[i|(1<<(j-1))] = min(f[i|(1<<(j-1))],f[i]);
g[i|(1<<(j-1))] = max(g[i|(1<<(j-1))],g[i]-a[j]);
} else{
f[i|(1<<(j-1))] = min(f[i|(1<<(j-1))],f[i]+1);
g[i|(1<<(j-1))] = max(g[i|(1<<(j-1))],w-a[j]);
}
} printf("%d",f[(1<<n)-1]); return 0;
}