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为了解决这个问题，我们需要找到一种方法，让多多在限定时间内尽可能多地采摘花生。多多只能从路边跳到相邻的花生植株，并采摘花生。每次移动和采摘都需要时间，我们需要优化采摘顺序以在限定时间内采摘最多的花生。

方法思路

解决代码

#include 
   
    #include 
    
     #include 
     
      using namespace std;int main() {    int M, N, K;    scanf("%d %d %d", &M, &N, &K);    vector
      
       
        > a(M + 1, vector
        
         (N + 1, 0)); for (int i = 1; i <= M; ++i) { int row_sum = 0; scanf("%d", &a[i][1]); for (int j = 2; j <= N; ++j) { int num; scanf("%d", &num); a[i][j] = row_sum + num; row_sum += num; } } vector
         
          
           > positions; vector
           
             nums; for (int i = 1; i <= M; ++i) { for (int j = 1; j <= N; ++j) { if (a[i][j] > 0) { positions.push_back(make_pair(i, j)); nums.push_back(a[i][j]); } } } sort(positions.begin(), positions.end(), [&nums](int i, int j) { return nums[i-1] > nums[j-1]; }); int ans = 0; int current_time = 0; int prev_i = 1; int prev_j = 1; for (int p = 0; p < positions.size(); ++p) { int i = positions[p].first; int j = positions[p].second; int num = nums[p]; int dx = abs(i - prev_i); int dy = abs(j - prev_j); int total = current_time + dx + dy + 1; if (total <= K) { ans += num; current_time = total; prev_i = i; prev_j = j; } } cout << ans << endl; return 0;}
           
          
         
        
       
      
     
    
   

代码解释