BZOJ3717 PA2014Pakowanie（状压dp）

　　显然贪心地有尽量先往容量大的背包里放。设f[i]为i子集物品最小占用背包数，g[i]为该情况下最后一个背包的剩余容量，转移显然。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 24
#define M 102
int n,m,a[N],b[M],f[1<<N],g[1<<N],lg2[1<<N];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3717.in","r",stdin);
    freopen("bzoj3717.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read(),m=read();
    for (int i=0;i<n;i++) a[i]=read();
    for (int i=1;i<=m;i++) b[i]=read();
    sort(b+1,b+m+1);reverse(b+1,b+m+1);
    for (int i=0;i<n;i++) lg2[1<<i]=i;
    memset(f,42,sizeof(f));
    f[0]=0;int s=(1<<n)-1;
    for (int i=0;i<(1<<n);i++)
    if (f[i]<=m)
        for (int j=s^i,t=j&-j;j;j^=t,t=j&-j)
        if (g[i]>=a[lg2[t]])
        {
            if (f[i]<f[i|t]) f[i|t]=f[i],g[i|t]=g[i]-a[lg2[t]];
            else if (f[i]==f[i|t]) g[i|t]=max(g[i|t],g[i]-a[lg2[t]]);
        }
        else
        if (b[f[i]+1]>=a[lg2[t]])
        {
            if (f[i]+1<f[i|t]) f[i|t]=f[i]+1,g[i|t]=b[f[i|t]]-a[lg2[t]];
            else if (f[i]+1==f[i|t]) g[i|t]=max(g[i|t],b[f[i|t]]-a[lg2[t]]);
        }
    if (f[(1<<n)-1]>m) cout<<"NIE";
    else cout<<f[(1<<n)-1];
    return 0;
}