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 = 1002;
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, 0, sizeof(dp)); 
        for (int i = 1; i <= n; i++)
            scanf("%d", &happy[i].Li);
        for (int i = 1; i <= n; i++)
            scanf("%d", &happy[i].Bi);
        sort(happy + 1, happy + n + 1, cmp);//贪心思想 
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= i && j <= v; j++)
                dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + happy[i].Li - happy[i].Bi * (j - 1));
        }
        printf("%d
", dp[n][v]);
    }
    
    return 0;
}

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 = 10002;
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, 0, sizeof(dp)); 
        for (int i = 1; i <= n; i++)
            scanf("%d", &happy[i].Li);
        for (int i = 1; i <= n; i++)
            scanf("%d", &happy[i].Bi);
        sort(happy + 1, happy + n + 1, cmp);//贪心思想 
        for (int i = 1; i <= n; i++)
        {
            for (int j = v; j >= 1; 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 - 1] + happy[i].Li - happy[i].Bi * (j - 1 ) > dp[j])
                    dp[j] = dp[j - 1] + happy[i].Li - happy[i].Bi * (j - 1 );
        }
        printf("%d
", dp[v]);
    }
    
    return 0;
}