hdoj 1171 Big Event in HDU（母函数/多重背包）

题意很简单，n种物品，每个物品对应一个价值和数量。让你竟可能的分成相等的两份。

可以设背包容量为sum/2用多重背包解决。

也可以用母函数来解决，计算能组成的最接近sum/2的值。

母函数可以解决很多背包问题。。但是速度一般比用dp做慢好多。。

母函数代码：1800ms

#include
using namespace std;
const int maxn = 1e5+5;
int v[55], m[55], n, c1[maxn], c2[maxn];
int main(void)
{
    while(cin >> n)
    {
        int sum = 0;
        if(n < 0) break;
        for(int i = 1; i <= n; i++)
            scanf("%d%d", &v[i], &m[i]), sum += v[i]*m[i];
        int mid = sum/2;
        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));
        for(int i = 0; i <= mid && i/v[1] <= m[1]; i += v[1])
            c1[i] = 1;
        for(int i = 2; i <= n; i++)
        {
            for(int j = 0; j <= mid; j++)
                for(int k = 0; k+j <= mid && k/v[i] <= m[i]; k += v[i])
                    c2[k+j] += c1[j];
            for(int j = 0; j <= mid; j++)
                c1[j] = c2[j], c2[j] = 0;
        }
        for(int i = mid; i >= 0; i--)
        {
            if(c1[i])
            {
                printf("%d %d", sum-i, i);
                break;
            }
        }
        printf("\n");
    }
    return 0;
}

多重背包代码：90ms

#include
using namespace std;
const int maxn = 1e5+5;
int dp[maxn], v[maxn], d[maxn], V[maxn];
int main(void)
{
    int n;
    while(cin >> n, n >= 0)
    {
        memset(dp, 0, sizeof(dp));
        int sum = 0;
        for(int i = 1; i <= n; i++)
            scanf("%d%d", &v[i], &d[i]), sum += v[i]*d[i];
        int mid = sum/2, count = 0;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= d[i]; j*=2)
            {
                V[count++] = j*v[i];
                d[i] -= j;
            }
            if(d[i] > 0) V[count++] = d[i]*v[i];
        }
        for(int i = 0; i < count; i++)
            for(int j = mid; j >= V[i]; j--)
                dp[j] = max(dp[j], dp[j-V[i]]+V[i]);
        printf("%d %d\n", sum-dp[mid], dp[mid]);
    }
    return 0;
}

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36879    Accepted Submission(s): 12814

Problem Description

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0

 

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0 A test case starting with a negative integer terminates input and this test case is not to be processed.

 

Output

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

 

Sample Input

2 10 1 20 1 3 10 1 20 2 30 1 -1

 

Sample Output

20 10 40 40

 

Author

lcy

 

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