二进制优化多重背包

二进制优化多重背包

$\bullet$题目：Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
$\bullet$Input
Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, …, n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0’’. The maximum total number of marbles will be 20000.

The last line of the input file will be ``0 0 0 0 0 0’’; do not process this line.

$\bullet$Output
For each colletcion, output Collection #k:'', where k is the number of the test case, and then eitherCan be divided.’’ or ``Can’t be divided.’’.

Output a blank line after each test case.
Sample Input

1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0

Sample Output

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

#include<cstdio>
#include<set>
#include<map>
#include<string.h>
#include<string>
#include<vector>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
int w[maxn],v[maxn],dp[maxn],num[maxn];
int  cnt=1;
void init(int x,int n)//二进制优化主要代码
{
    int num=1;
    while(n>num){
        w[cnt]=num*x;
        v[cnt++]=num*x;
        n-=num;
        num<<=1;
    }
    w[cnt]=n*x;
    v[cnt++]=n*x;
}
int main()
{
    int a[7],x=1;
    while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])&&a[1]+a[2]+a[3]+a[4]+a[5]+a[6])
    {
        memset(dp,0,sizeof(dp));
        int sum=0;
        cnt=1;
        for(int i=1;i<=6;i++){
            sum+=a[i]*i;
            init(i,a[i]);
        }
        cnt--;
        if(sum%2==1){
            printf("Collection #%d:\nCan't be divided.\n",x++);
        }
        else {
            int W=sum/2;
            for(int i=1;i<=cnt;i++)
            for(int j=W;j>=w[i];j--)
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            int temp=sum-dp[W];
            if(temp==dp[W]){
                printf("Collection #%d:\nCan be divided.\n",x++);
            }
            else {
                printf("Collection #%d:\nCan't be divided.\n",x++);
            }
        }
        cout<<endl;
    }
    return 0;
}

再来一题：
http://acm.hdu.edu.cn/showproblem.php?pid=1171
二进制优化后可以大大减少时间复杂度

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<iostream>
#include<queue>
#include<string>
#include<math.h>
typedef long long LL;
using namespace std;
const int maxn=1e5+10;
int n,cnt=0,v[maxn],dp[maxn],w[maxn];

void add(int x,int n)//二进制优化
{
    int num = 1;
    while(num < n){
        v[++cnt] = num * x;
        w[cnt] = num * x;
        n -= num;
        num *= 2;
    }
    v[++cnt] = n * x;
    w[cnt] = n * x;
}

void init()
{
    cnt = 0;
    memset(v,0,sizeof(v));
    memset(w,0,sizeof(w));
    memset(dp,0,sizeof(dp));
}

int main()
{
    while(scanf("%d",&n)&&n>0)
    {
        init();
        int sum = 0;
        for(int i=1;i<=n;i++){
            int w,num;
            scanf("%d%d",&w,&num);
            add(w,num);
            sum += w * num;
        }
        int V = sum / 2;

        for(int i = 1 ; i <= cnt ; i++){
            for(int j = V ; j >= w[i] ; j--){
                dp[j] = max(dp[j] , dp[j - w[i]] + v[i]);
            }
        }

        printf("%d %d\n",max(dp[V],sum - dp[V]),min(dp[V],sum - dp[V]));
    }
    return 0;
}