【日常學習】【多重背包】【二進制優化】hdu1059 Dividing題解

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  

堪稱多重背包的模闆題了，要用二進制優化，比較函數要手寫

題意就不說了，網上搜吧 

我們做一個容量為總價值的一半的背包即可。

注意一定要一開始f[0]=0,其他都是-oo  雖然一開始都指派為0最後檢驗f[v]==v理論上也可行，但是目測資料有問題，會莫名RE

還是永福處置保證隻有裝滿才是合法解吧，保險。

代碼：

//hdu1059 Dividing 多重背包練習 二進制優化
//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)//體積，價值 
{
	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這幾天一個Open Judge擾得我心煩意亂，一定要冷靜

——昨夜風兼雨，簾帏飒飒秋聲。燭殘漏斷頻欹枕，起坐不能平。