hihoCoder#1044 状态压缩·一

原题地址

 

跟背包问题非常像，很巧妙，跟着提示做即可

需要注意的是，状态压缩以后，j的变化不是规律的，所以用一个临时缓冲back保存下一次迭代的结果。

 

代码：

 1 #include <iostream>

 2 

 3 using namespace std;

 4 

 5 int onesum(int a) {

 6   int sum = 0;

 7   while (a) {

 8     a &= (a - 1);

 9     sum++;

10   }

11   return sum;

12 }

13 

14 int main() {

15   int N, M, Q;

16   int w[1024] = {0};

17   int res[1024] = {0};

18   int back[1024] = {0};

19   int mask = 0;

20 

21   cin >> N >> M >> Q;

22   for (int i = 0; i < N; i++)

23     cin >> w[i];

24 

25   mask = (1 << (M - 1)) - 1;

26   for (int i = N - 1; i >= 0; i--) {

27     for (int j = 0; j <= mask; j++) {

28       if (onesum(j) < Q)

29         back[j] = max(res[((j << 1) + 1) & mask] + w[i], res[(j << 1) & mask]);

30       else

31         back[j] = res[(j << 1) & mask];

32     }

33     for (int j = 0; j <= mask; j++)

34       res[j] = back[j];

35   }

36 

37   cout << res[0] << endl;

38   return 0;

39 }