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)：

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <string.h>
 4 #include <algorithm>
 5 using namespace std;
 6 struct Happy{
 7     int Li, Bi;
 8 };
 9 const int N = 1002;
10 int dp[N][N];
11 Happy happy[N];
12 bool cmp(Happy a, Happy b)//递减排序
13 {
14     return a.Bi > b.Bi;
15 }
16 int main()
17 {
18     int n, v;
19     while (~scanf("%d %d", &n, &v))
20     {
21         memset(dp, 0, sizeof(dp));
22         for (int i = 1; i <= n; i++)
23             scanf("%d", &happy[i].Li);
24         for (int i = 1; i <= n; i++)
25             scanf("%d", &happy[i].Bi);
26         sort(happy + 1, happy + n + 1, cmp);//贪心思想
27         for (int i = 1; i <= n; i++)
28         {
29             for (int j = 1; j <= i && j <= v; j++)
30                 dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + happy[i].Li - happy[i].Bi * (j - 1));
31         }
32         printf("%d\n", dp[n][v]);
33     }
34
35     return 0;
36 }

代码二(优化空间版)：

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <string.h>
 4 #include <algorithm>
 5 using namespace std;
 6 struct Happy{
 7     int Li, Bi;
 8 };
 9 const int N = 10002;
10 int dp[N];
11 Happy happy[N];
12 bool cmp(Happy a, Happy b)//递减排序
13 {
14     return a.Bi > b.Bi;
15 }
16 int main()
17 {
18     int n, v;
19     while (~scanf("%d %d", &n, &v))
20     {
21         memset(dp, 0, sizeof(dp));
22         for (int i = 1; i <= n; i++)
23             scanf("%d", &happy[i].Li);
24         for (int i = 1; i <= n; i++)
25             scanf("%d", &happy[i].Bi);
26         sort(happy + 1, happy + n + 1, cmp);//贪心思想
27         for (int i = 1; i <= n; i++)
28         {
29             for (int j = v; j >= 1; j--)
30             /*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*/
31                 if (dp[j - 1] + happy[i].Li - happy[i].Bi * (j - 1 ) > dp[j])
32                     dp[j] = dp[j - 1] + happy[i].Li - happy[i].Bi * (j - 1 );
33         }
34         printf("%d\n", dp[v]);
35     }
36
37     return 0;
38 }