【日常学习】【多重背包】【二进制优化】hdu1059 Dividing题解

左手的ㄟ右手 2022-08-09 19:29 191阅读 0赞

# Dividing #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 21177    Accepted Submission(s): 5976**

Problem Description

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.

Source

[Mid-Central European Regional Contest 1999][]

堪称多重背包的模板题了，要用二进制优化，比较函数要手写

题意就不说了，网上搜吧

我们做一个容量为总价值的一半的背包即可。

注意一定要一开始f\[0\]=0,其他都是-oo  虽然一开始都赋值为0最后检验f\[v\]==v理论上也可行，但是目测数据有问题，会莫名RE

还是永福处置保证只有装满才是合法解吧，保险。

代码：

//hdu1059 Dividing 多重背包练习 二进制优化
    //copyright by ametake
    #include
       
        
    #include
        
         
    #include
         
          
    using namespace std;
    
    const int maxv=12000000+10;
    const int oo=0x3f3f3f3f;
    
    int a[10],p=0;
    int v,f[maxv];
    
    void zerooneback(int c,int w)//体积，价值 
    {
    	for (int i=v;i>=c;i--){if (f[i]
          
           =v)
    	{
    		completepack(c,w);
    		return;
    	}
    	int k=1;
    	while (k
           
            =0) printf("Collection #%d:\nCan be divided.\n\n",++p); else printf("Collection #%d:\nCan't be divided.\n\n",++p); } return 0; }

CCF这几天一个Open Judge扰得我心烦意乱，一定要冷静

——昨夜风兼雨，帘帏飒飒秋声。烛残漏断频欹枕，起坐不能平。

[Mid-Central European Regional Contest 1999]: http://acm.hdu.edu.cn/search.php?field=problem&key=Mid-Central+European+Regional+Contest+1999&source=1&searchmode=source