hdu 1059 Dividing（二进制转化优化） - - ITeye技术网站

hdu 1059 Dividing（二进制转化优化） - - ITeye技术网站

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.

 两个人共同收藏了一些石头，现在要分道扬镳，得分资产了，石头具有不同的收藏价值，分别为1、2、3、4、5、6共6个价钱。问：是否能公平分配？

输入：每行为一个测试例子，每行包括6个数字，分别对应6种价钱的石头数目，比如101200代表价值为1的石头有1个，价值为2的石头有0个....价值为4的石头有2个。他们具有的石头数量的上限为2万个。

        此题属于多重背包，数据很大，需要二进制优化，不然超时。优化后直接当做01背包来做，01背包是逆序！

1. int bag[120005];  
2. int w[15], n[15], v[1005];  
3.   
4. void _01_bag()  
5. {  
6.     int i, j;  
7.     memset(bag, 0, sizeof(bag));  
8.     for(i = 0; i < total; i++)  
9.     {  
10.         for(j = sum; j >= v[i]; j--)  
11.         {  
12.             bag[j] = max(bag[j], bag[j-v[i]]+v[i]);  
13.         }  
14.     }  
15. }  
16.   
17. int main()  
18. {  
19.     int i, temp, zz = 1;  
20.     while(scanf("%d %d %d %d %d %d", &n[0], &n[1], &n[2], &n[3], &n[4], &n[5]))  
21.     {  
22.         if(n[0] + n[1] + n[2] + n[3] + n[4] + n[5] == 0) break;  
23.         printf("Collection #%d:\n", zz++);  
24.         V = 0;  
25.         for(i = 0; i < N; i++)  
26.         {  
27.             w[i] = i + 1;  
28.             V += w[i]*n[i]; //求总和  
29.         }  
30.         if(V%2 == 1)    //总和是奇数则不能平分  
31.         {  
32.             printf("Can't be divided.\n\n");  
33.             continue;  
34.         }  
35.         sum = V/2; total = 0;  
36.         for(i = 0; i < N; i++)  //二进制压缩为——01背包  
37.         {  
38.             if(n[i] == 0) continue;  
39.             temp = 1;  
40.             while(n[i] > temp)  
41.             {  
42.                 v[total++] = temp*w[i]; //将新的值赋给v[]  
43.                 n[i] -= temp;  
44.                 temp *= 2;  
45.             }  
46.             v[total++] = n[i]*w[i];  
47.         }  
48.         _01_bag();  //用新的v[]数组直接拿来01背包  
49.         if(bag[sum] != sum)  
50.             printf("Can't be divided.\n\n");  
51.         else  
52.             printf("Can be divided.\n\n");  
53.     }  
54.   
55.     return 0;  
56. } 

http://blog.csdn.net/u013081425/article/details/19838357

1. //优化的多重背包  
2. void solve()  
3. {  
4.     sum = sum>>1;  
5.     memset(dp,0,sizeof(dp));  
6.     dp[0] = 1;  
7.   
8.     for(int i = 1; i <= 6; i++)  
9.     {  
10.         if(num[i] > 0)  
11.         {  
12.             memset(use,0,sizeof(use));  
13.             for(int j = i; j <= sum; j++)  
14.             {  
15.                 if(dp[j-i] && !dp[j] && use[j-i] < num[i])  
16.                 {  
17.                     dp[j] = 1;  
18.                     use[j] = use[j-i]+1;  
19.                 }  
20.             }  
21.         }  
22.     }  
23.   
24.     if(dp[sum]) printf("Can be divided.\n\n");  
25.     else printf("Can't be divided.\n\n");  
26. } 

http://www.cnblogs.com/andyqsmart/p/4517673.html

void Work()
{
    if (sum % 2)
    {
        printf("Can't be divided.\n\n");
        return;
    }
    sum /= 2;
    dp[0] = true;
    for (int i = 1; i <= 6; ++i)
    {
        if (num[i] == 0)
            continue;
        memset(vis, 0, sizeof(vis));
        for (int j = i; j <= sum; j++)
        {
            if (dp[j-i] && !dp[j] && vis[j-i] < num[i])
            {
                dp[j] = true;
                vis[j] = vis[j-i]+1;
            }
        }
    }
    if (dp[sum])
        printf("Can be divided.\n\n");
    else
        printf("Can't be divided.\n\n");
}

http://www.faceye.net/search/92019.html

15     while(scanf("%d %d %d %d %d %d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])==6&&(a[1]+a[2]+a[3]+a[4]+a[5]+a[6])){
16         n=0;
17     memset(dp,0,sizeof(dp));
18         printf("Collection #%d:\n",kase++);
19         sum=a[1]+a[2]*2+a[3]*3+a[4]*4+a[5]*5+a[6]*6;
20         if(sum&1){
21             printf("Can't be divided.\n\n");continue;
22         }
23         for(i=1;i<=6;i++){
24             n=0;
25             for(j=1;j<=a[i];j<<=1){
26                 v[n++]=j;
27                 a[i]-=j;
28             }
29             if(a[i]>0){
30                 v[n++]=a[i];
31             }
32             for(j=0;j<n;j++){
33             for(k=sum/2;k>=v[j]*i;k--)
34             dp[k]=max(dp[k],dp[k-v[j]*i]+v[j]*i);
35           }
36         }
37         
38         if(dp[sum/2]==sum/2){
39             printf("Can be divided.\n\n");
40         }
41         else printf("Can't be divided.\n\n");

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