BZOJ 1190 梦幻岛宝珠(分组01背包)

跑了7000ms。。。

这是个体积和价值都超大的背包。但是体积保证为a*2^b的(a<=10,b<=30)形式.且n<=100.

于是可以想到按b来分组。这样的话每组最多为a*n*2^b的体积。把每个分组进行一次01背包。

令dp[i][j]表示体积为j*2^i+w&(1<<i)的最大价值。再把每个分组背包做一次01背包，随便转移一下就行了。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 1000000000
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
//# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=105;
//Code begin...

int dp[35][1005];

int main ()
{
    int n, W, x, val, wei, tmp;
    while (scanf("%d%d",&n,&W)) {
        if (n==-1) break;
        tmp=W; wei=0; mem(dp,0);
        while (tmp) ++wei, tmp>>=1;
        FOR(i,1,n) {
            scanf("%d%d",&x,&val);
            int num=0;
            while (x) {
                ++num;
                if (x&1) break;
                x>>=1;
            }
            for (int i=1000; i>=x; --i) dp[num][i]=max(dp[num][i],dp[num][i-x]+val);
        }
        FOR(i,1,wei) for (int j=1000; j>=0; --j) for (int k=j; k>=0; --k) {
            int l=min(((j-k)<<1)+((W>>(i-2))&1),1000);
            dp[i][j]=max(dp[i][j],dp[i][k]+dp[i-1][l]);
        }
        printf("%d\n",dp[wei][1]);
    }
    return 0;
}