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 = ;
 const int M =  * ;
 int v[N];
 int dp[M];

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

         memset(dp, , sizeof(dp));
         for (int i = ; i < cur; i++)
         {
             for (int j = sum / ; 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 / ] << " " << dp[sum / ] << endl;
     }
     return ;
 

多重背包：

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

 const int N = ;
 const int M =  * ;
 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 = ;
         while (k <= amount)
         {
             zero_one_pack(weight*k, value*k, capacity);
             amount -= k;
             k *= ;
         }
         zero_one_pack(weight*amount, value*amount, capacity);
     }

 }

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

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

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