M - Dividing(完全背包+二进制优化)

偏执的太偏执、 2022-04-23 16:06 179阅读 0赞

### 题意： ###

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.  
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.  
Output For each colletcion, output Collection \#k:’’, where k is the number of the test case, and then either\`Can 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.

##### 题意： #####

给出1到6的价值的物品个数。 问你是否能把他们平分。

##### 分析： #####

是个完全背包，但是必须二进制优化才能过。

#include"stdio.h"
    #include"string.h"
    #include"functional"
    #include"iostream"
    #include"algorithm"
       using namespace std;
    typedef long long ll;
    #define scanll(a,b) scanf("%I64d%I64d",&a,&b);
    #define scanl(a) scanf("%I64d",&a);
    #define scanff(a,b) scanf("%lf%lf",&a,&b);
    #define scan1f(a) scanf("%lf",&a);
    #define prinll(a,b) printf("%I64d %I64d",a,b);
    #define prinl(a) printf("%I64d",a);
    #define printff(a,b) printf("%lf %lf",a,b);
    #define printlf(a) printf("%lf",a);
    #define OK printf("\n");
    #define MAXSIXE 310
    ll nc[200010];
    ll nw[200010];
    int main()
    {
        ll a[7];
        ll cnt=1;
        while(1)
        {
            ll sum=0;
            ll s=0;
            for(ll i=1;i<=6;i++)
            {
                scanl((a[i])); sum+=a[i];s+=a[i]*i;
            }
            if(sum==0)  break;  sum=s;
            if(sum%2==1)//奇数不可能平分
            {
                printf("Collection #%lld:\n",cnt++);
                printf("Can't be divided.\n"); OK; continue;
            }
            ll dp[200010];
            ll ncnt=1;
            memset(dp,0,sizeof(dp));
            for(ll i=1;i<=6;i++)//二进制优化
            {
                ll k;
                for(k=1;k <= a[i]; k <<= 1)
                {
                    nc[ncnt] = k * i;
                    nw[ncnt] = k * i;
                    ncnt++;
                    a[i] -= k;
                }
                if( a[i] > 0)
                {
                    nc[ncnt] = a[i] * i;
                    nw[ncnt] = a[i] * i;
                    ncnt++;
                }
            }
            ll MARK = 0;//01背包跑一遍 
            for(ll i = 1;i < ncnt; i++)
            {
                for(ll j = sum/2; j >= nw[i]; j--)
                {
                    dp[j] = max(dp[j],dp[j-nw[i]] + nc[i]);
                }
            }
            if(dp[sum/2] != sum/2)
            {
                printf("Collection #%lld:\n",cnt++);
                printf("Can't be divided.\n"); OK; continue;
            }
            else
            {
                printf("Collection #%lld:\n",cnt++);
                printf("Can be divided.\n"); OK; continue;
            }
    
        }
        return 0;
    }