1014 Dividing dp/dfs

先用dfs方法:(注意剪枝的重要性)

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int sumvalue,halfvalue;
int n[7];
bool flag;//记录是否能够进行平分。
void dfs(int value,int c){
    if(flag)
        return;
    if(value==halfvalue){
        flag=true;
        return;
    }
    for(int i=c;i>=1;i--){
        if(n[i]){
            if(value+i<=halfvalue){//注意<=号的意义
                n[i]--;
                dfs(value+i,i);
                if(flag)
                    break;
            }
        }
    }
    return ;
}
int main(){
    int k=0;
    while(true){
        sumvalue=0;//注意sumvalue初始化的位置,而且必须进行初始化。
        k++;
    for(int i=1;i<=6;i++){
        cin>>n[i];
        sumvalue+=(i*n[i]);
    }
    if(sumvalue==0)
        break;
    if(sumvalue%2!=0){//为真
        cout<<"Collection #"<<k<<":"<<endl;
        cout<<"Can't be divided."<<endl;
        cout<<endl;
        continue;
    }
    halfvalue=sumvalue/2;
    flag=false;//为了验证dfs的效果。
    dfs(0,6);
    if(flag){
        cout<<"Collection #"<<k<<':'<<endl;
        cout<<"Can be divided."<<endl<<endl;
        continue;
        }
    else{
        cout<<"Collection #"<<k<<':'<<endl;
        cout<<"Can't be divided."<<endl<<endl;
        continue;
        }
    }
    return 0;
}

多重背包+二进制优化

来源:https://blog.csdn.net/lyy289065406/article/details/6661449

#include<iostream>
using namespace std;
 
int n[7];  //价值为i的物品的个数
int v;  //背包容量
int SumValue;  //物品总价值
bool flag;    //标记是否能平分SumValue
int dp[100000];  //状态数组
 
int max(int a,int b)
{
    return a>b?a:b;
}
 
/*完全背包*/
void CompletePack(int cost,int weight)
{
    for(int i=cost;i<=v;i++)
    {
        dp[i]=max(dp[i],dp[i-cost]+weight);
        if(dp[i]==v)    //剪枝,当能够平分SumValue时退出
        {
            flag=true;
            return;
        }
    }
            
    return;
}
 
/*01背包*/
void ZeroOnePack(int cost,int weight)
{
    for(int i=v;i>=cost;i--)
    {
        dp[i]=max(dp[i],dp[i-cost]+weight);
        if(dp[i]==v)    //剪枝
        {
            flag=true;
            return;
        }
    }
    return;
}
 
/*多重背包*/
void MultiplePack(int cost,int weight,int amount)
{
    if(cost*amount>=v)
    {
        CompletePack(cost,weight);
        return;
    }
 
    if(flag)    //剪枝
        return;
 
    /*二进制优化*/
    int k=1;
    while(k<amount)
    {
        ZeroOnePack(k*cost,k*weight);
 
        if(flag)    //剪枝
            return;
 
        amount-=k;
        k*=2;
    }
    ZeroOnePack(amount*cost,amount*weight);
 
    return;
}
 
int main(int i)
{
    int test=1;
    while(cin>>n[1]>>n[2]>>n[3]>>n[4]>>n[5]>>n[6])
    {
        SumValue=0;  //物品总价值
 
        for(i=1;i<=6;i++)
            SumValue+=i*n[i];
 
        if(SumValue==0)
            break;
 
        if(SumValue%2)    //sum为奇数,无法平分
        {
            cout<<"Collection #"<<test++<<':'<<endl;
            cout<<"Can't be divided."<<endl<<endl;    //注意有空行
            continue;
        }
 
        v=SumValue/2;
        memset(dp,-1,sizeof(dp));
        dp[0]=0;
        flag=false;
 
        for(i=1;i<=6;i++)
        {
            MultiplePack(i,i,n[i]);
 
            if(flag)    //剪枝
                break;
        }
 
        if(flag)
        {
            cout<<"Collection #"<<test++<<':'<<endl;
            cout<<"Can be divided."<<endl<<endl;
            continue;
        }
        else
        {
            cout<<"Collection #"<<test++<<':'<<endl;
            cout<<"Can't be divided."<<endl<<endl;
            continue;
        }
    }
    return 0;
}

 

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转载自www.cnblogs.com/sweet-ginger-candy/p/11517257.html