HDU -2670 Girl Love Value

这道题是刚好装满的背包问题，刚好选取k个，状态转移方程为dp[i][j] = max( dp[i - 1][j], dp[i - 1][j - 1] + Li - Bi(j - 1) )

dp[i][j] 表示从前 i 个男孩中选取 j 个的 Li 的最大值， 首先按照Bi 排一下序，这个是利用贪心的思想，因为那样才能获得最优解，按照递减排序，这样才能找到最大值，然后就是dp了，状态转移方程的意思就是第 j 个去或者不 取，代码如下

代码一(二维数组版)：

 #include <stdio.h>
 #include <iostream>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 struct Happy{
     int Li, Bi;
 };
 const int N = ;
 int dp[N][N];
 Happy happy[N];
 bool cmp(Happy a, Happy b)//递减排序
 {
     return a.Bi > b.Bi;
 }
 int main()
 {
     int n, v;
     while (~scanf("%d %d", &n, &v))
     {
         memset(dp, , sizeof(dp));
         for (int i = ; i <= n; i++)
             scanf("%d", &happy[i].Li);
         for (int i = ; i <= n; i++)
             scanf("%d", &happy[i].Bi);
         sort(happy + , happy + n + , cmp);//贪心思想
         for (int i = ; i <= n; i++)
         {
             for (int j = ; j <= i && j <= v; j++)
                 dp[i][j] = max(dp[i - ][j], dp[i - ][j - ] + happy[i].Li - happy[i].Bi * (j - ));
         }
         printf("%d\n", dp[n][v]);
     }

     return ;
 }

代码二(优化空间版)：

 #include <stdio.h>
 #include <iostream>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 struct Happy{
     int Li, Bi;
 };
 const int N = ;
 int dp[N];
 Happy happy[N];
 bool cmp(Happy a, Happy b)//递减排序
 {
     return a.Bi > b.Bi;
 }
 int main()
 {
     int n, v;
     while (~scanf("%d %d", &n, &v))
     {
         memset(dp, , sizeof(dp));
         for (int i = ; i <= n; i++)
             scanf("%d", &happy[i].Li);
         for (int i = ; i <= n; i++)
             scanf("%d", &happy[i].Bi);
         sort(happy + , happy + n + , cmp);//贪心思想
         for (int i = ; i <= n; i++)
         {
             for (int j = v; j >= ; j--)
             /*这句话等价于dp[j] = max(dp[j], dp[j - 1] + happy[i].Li - happy[i].Bi * (j - 1))，只不过用if更快*/
                 if (dp[j - ] + happy[i].Li - happy[i].Bi * (j -  ) > dp[j])
                     dp[j] = dp[j - ] + happy[i].Li - happy[i].Bi * (j -  );
         }
         printf("%d\n", dp[v]);
     }

     return ;
 }