hdoj1171 Big Event in HDU（01背包 || 多重背包）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1171

题意

老师有一个属性：价值（value）。在学院里的老师共有n种价值，每一种价值value对应着m个老师，说明这m个老师的价值都为value。现在要将这些老师从人数上平分成两个院系，并且希望平分后两个院系老师的总价值A和B应尽可能地相等，求A和B的值（A>=B）。

思路

由于每种老师的个数是有限的，所以使用多重背包解决。由于测试数据不是很严格，所以使用01背包也可以通过。

代码

01背包：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 5010;
const int M = 50 * 5000;
int v[N];
int dp[M];

int main()
{
    //freopen("hdoj1171.txt", "r", stdin);
    int n;
    while (cin >> n && n >= 0)
    {
        int cur = 0;    //记录教师总数
        int sum = 0;    //记录教师总价值
        for (int i = 0;i < n; i++)
        {
            int val, m;
            cin >> val >> m;
            for (int j = 0; j < m; j++)
            {
                v[cur++] = val;
                sum += val;
            }
        }

        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < cur; i++)
        {
            for (int j = sum / 2; j >= v[i]; j--)
                dp[j] = max(dp[j], dp[j - v[i]] + v[i]);    //weight和value相同
        }
        //由于dp[sum/2]<sum/2,所以dp[sum/2]一定是较小者，sum - dp[sum / 2]是较大者
        cout << sum - dp[sum / 2] << " " << dp[sum / 2] << endl;
    }
    return 0;

多重背包：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 110;
const int M = 50 * 5000;
int v[N], num[N];
int dp[M];
int sum;

void zero_one_pack(int weight, int value, int capacity)
{
    for (int i = capacity; i >= weight; i--)    //逆序
        dp[i] = max(dp[i], dp[i - weight] + value);
}

void complete_pack(int weight, int value, int capacity)
{
    for (int i = weight; i <= capacity; i++)    //正序
        dp[i] = max(dp[i], dp[i - weight] + value);
}

void mutiple_pack(int weight, int value, int amount, int capacity)
{
    if (weight*amount >= capacity)
        complete_pack(weight, value, capacity);
    else
    {
        int k = 1;
        while (k <= amount)
        {
            zero_one_pack(weight*k, value*k, capacity);
            amount -= k;
            k *= 2;
        }
        zero_one_pack(weight*amount, value*amount, capacity);
    }

}

int main()
{
    //freopen("hdoj1171.txt", "r", stdin);
    int n;
    while (cin >> n && n >= 0)
    {
        sum = 0;
        for (int i = 1; i <= n; i++)
        {
            cin >> v[i] >> num[i];
            sum += v[i] * num[i];
        }

        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; i++)
            mutiple_pack(v[i], v[i], num[i], sum / 2);

        cout << sum - dp[sum / 2] << " " << dp[sum / 2] << endl;
    }
}