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：

#include <iostream>

using namespace std;

int onesum(int a) {
  int sum = 0;
  while (a) {
    a &= (a - 1);
    sum++;
  }
  return sum;
}

int main() {
  int N, M, Q;
  int w[1024] = {0};
  int res[1024] = {0};
  int back[1024] = {0};
  int mask = 0;

  cin >> N >> M >> Q;
  for (int i = 0; i < N; i++)
    cin >> w[i];

  mask = (1 << (M - 1)) - 1;
  for (int i = N - 1; i >= 0; i--) {
    for (int j = 0; j <= mask; j++) {
      if (onesum(j) < Q)
        back[j] = max(res[((j << 1) + 1) & mask] + w[i], res[(j << 1) & mask]);
      else
        back[j] = res[(j << 1) & mask];
    }
    for (int j = 0; j <= mask; j++)
      res[j] = back[j];
  }

  cout << res[0] << endl;
  return 0;
}