[Algorithm][综合训练][比那名居的桃子][chika和蜜柑][礼物的最大价值]详细讲解

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1.比那名居的桃子

1.题目链接

• 比那名居的桃子

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2.算法原理详解 && 代码实现

• 自己的版本：暴力解法 --> 超时 37.5%

  #include 
  #include 
  using namespace std;
  
  int main()
  {
      int n = 0, k = 0;
      cin >> n >> k;
  
      vector<int> happy(n, 0), shame(n, 0);
      for(int i = 0; i < n; i++)
      {
          cin >> happy[i];
      }
  
      for(int i = 0; i < n; i++)
      {
          cin >> shame[i];
      }
  
      vector<vector<int>> ret(n, vector<int>(2, 0));
      for(int i = 0; i < n; i++) // 枚举每个起点
      {
          for(int j = 0; j < k; j++)
          {
              if(i + j < n)
              {
                  ret[i][0] += happy[i + j];
                  ret[i][1] += shame[i + j];
              }
          }
      }
  
      int day = 0, happyMax = 0;
      for(int i = 0; i < n; i++)
      {
          // 目标是：尽可能多的快乐值
          if(happyMax < ret[i][0])
          {
              happyMax = ret[i][0];
              day = i;
          }
  
          // 快乐值相等 -->  尽可能少的羞耻度
          if(happyMax == ret[i][0])
          {
              if(ret[i][1] < ret[day][1])
              {
                  day = i;
              }
          }
  
          // 快乐值 == 羞耻度 -->  尽可能早的吃桃子
          // 不更新值就可以
      }
  
  
      cout << day + 1<< endl;
  
      return 0;
  }
  

• 优化版本1：滑动窗口

  #include 
  #include 
  using namespace std;
  
  int main()
  {
      int n = 0, k = 0;
      cin >> n >> k;
  
      vector<int> happy(n, 0), shame(n, 0);
      for(int i = 0; i < n; i++)
      {
          cin >> happy[i];
      }
  
      for(int i = 0; i < n; i++)
      {
          cin >> shame[i];
      }
  
      int left = 0, right = 0;
      long long hSum = 0, sSum = 0, hMax = 0, sMin = 0, begin = 0;
      
  	while(right < n)
      {
  		hSum += happy[right];
          sSum += shame[right];
          
          while(right - left + 1 > k)
          {
              hSum -= happy[left];
              sSum -= shame[left];
              left++;
          }
          
          if(right - left + 1 == k)
          {
              if(hSum > hMax)
              {
                  begin = left;
                  hMax = hSum;
                  sMin = sSum;
              }
              else if(hSum == hMax && sSum < sMin)
              {                
                  begin = left;
                  sMin = sSum;
              }
          }
          
          right++;
      }
  
      cout << begin + 1<< endl;
  
      return 0;
  }
  

• 优化版本2：前缀和

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2.chika和蜜柑

1.题目链接

• chika和蜜柑

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2.算法原理详解 && 详细讲解

• 优化版本：排序/TOP-K --> 堆/数组模拟均可 --> 自定排序规则

  #include 
  #include 
  #include 
  using namespace std;
  
  typedef pair<int, int> PII;
  
  int main()
  {
  	int n = 0, k = 0;
  	cin >> n >> k;
  
  	vector<PII> orgs(n); // 
  	for(int i = 0; i < n; i++)
  	{
  		cin >> orgs[i].first;
  	}
  
  	for(int i = 0; i < n; i++)
  	{
  		cin >> orgs[i].second;
  	}
  
  	sort(orgs.begin(), orgs.end(), [&](const PII& p1, const PII& p2)
  	{
  		if(p1.second != p2.second)
  		{
  			return p1.second > p2.second;
  		}
  		else
  		{
  			return p1.first < p2.first;
  		}
  
  	});
  
  	long long sour = 0, sweet = 0;
  	for(int i = 0; i < k; i++)
  	{
  		sour += orgs[i].first;
  		sweet += orgs[i].second;
  	}
  
  	cout << sour << " " << sweet << endl;
  
  	return 0;
  }
  

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3.礼物的最大价值

1.题目链接

• 礼物的最大价值

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2.算法原理详解 && 代码实现

• 自己的版本：暴搜 --> 超时

  class Solution 
  {
  public:
      int dx[2] = {1, 0}; 
      int dy[2] = {0, 1};
      int n = 0, m = 0, cnt = 0, ret = 0;
  
      int maxValue(vector<vector<int> >& grid) 
      {
          n = grid.size(), m = grid[0].size();
          DFS(grid, grid[0][0], 0, 0);
  
          return ret;
      }
  
      void DFS(vector<vector<int> >& grid, int cnt, int x, int y)
      {
          if(x == n - 1 && y == m - 1)
          {
              ret = max(cnt, ret);
          }
  
          for(int k = 0; k < 2; k++)
          {
              int a = x + dx[k], b = y + dy[k];
              if(a < n && b < m)
              {
                  DFS(grid, cnt + grid[a][b], a, b);
              }
          }
      }
  };
  

• 优化版本：动态规划 – 路径问题

  int maxValue(vector<vector<int> >& grid)
  {
  	int n = grid.size(), m = grid[0].size();
  	vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
  
  	for(int i = 1; i <=n; i++)
  	{
  		for(int j = 1; j <= m; j++)
  		{
  			 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + grid[i - 1][j - 1];
  		}
  	}
  
  	return dp[n][m];
  }