problem1

 计算每个格子向上的最大高度。然后每个格子同一行前面的格子以及当前格子作为选取的矩形的最后一行，计算面积并更新答案。

problem2

对于两个数据$(x_{1},y_{1}),(x_{2},y_{2})$,若先完成第一个再完成第二个，那么一开始的值$F$需要满足$F\geq max(x_{1}, x_{2}+(x_{1}-y_{1}))$,反过来需要满足$F\geq max(x_{2}, x_{1}+(x_{2}-y_{2}))$。所以若前者更优的话，那么有$max(x_{1}, x_{2}+(x_{1}-y_{1}))<max(x_{2}, x_{1}+(x_{2}-y_{2}))$

由于$x_{1}>y_{1}, x_{2}>y_{2}$，所以等价于$y_{1}<y_{2}$。所以按照$y$升序排序，然后从前向后dp即可。

problem3

最后最优值跟$x$的函数关系是多个线段，且这些线段是一个凸函数。如下图的棕色线所示。从后向前扩展每个点。每次扩展相当于把之前的折线从最高处垂直分开然后向两边平移一段距离，然后加上当前点的代价。

 

code for problem1

#include 
     
      #include 
      
       class TheMatrix { public:  int MaxArea(const std::vector
       
         &board) {    int n = static_cast
        
         (board.size());    int m = static_cast
         
          (board[0].size());    std::vector
          
           
            > h(n, std::vector
            
             (m)); int result = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { if (i == 0 || board[i][j] == board[i - 1][j]) { h[i][j] = 1; } else { h[i][j] = h[i - 1][j] + 1; } result = std::max(result, h[i][j]); int min_h = h[i][j]; for (int k = j - 1; k >= 0 && board[i][k] != board[i][k + 1]; --k) { min_h = std::min(min_h, h[i][k]); result = std::max(result, min_h * (j - k + 1)); } } } return result; }};
            
           
          
         
        
       
      
     

code for problem2

#include 
     
      #include 
      
       #include 
       
        class AlbertoTheAviator { public:  int MaximumFlights(int F, const std::vector
        
          &duration,                     const std::vector
         
           &refuel) {    int n = static_cast
          
           (duration.size()); std::vector
           
             indices(n); for (int i = 0; i < n; ++i) { indices[i] = i; } std::sort(indices.begin(), indices.end(), [&](int l, int r) { return refuel[l] > refuel[r]; }); std::vector
            
             
              > f(n, std::vector
              
               (F + 1)); for (int i = duration[indices[n - 1]]; i <= F; ++i) { f[n - 1][i] = 1; } for (int i = n - 2; i >= 0; --i) { for (int j = 1; j <= F; ++j) { f[i][j] = f[i + 1][j]; if (j >= duration[indices[i]]) { f[i][j] = std::max( f[i][j], 1 + f[i + 1][j - duration[indices[i]] + refuel[indices[i]]]); } } } return f[0][F]; }};
              
             
            
           
          
         
        
       
      
     

code for problem3

import java.math.*;import java.util.*;public class MiningGoldHard {  public int GetMaximumGold(int n, int m, int[] event_i, int[] event_j, int[] event_di, int[] event_dj) {    return Solve(n, event_i, event_di) + Solve(m, event_j, event_dj);  }  int Solve(int N, int[] e, int[] d) {    int m = e.length;    List
     
       ends = new ArrayList
      
       ();    ends.add(new Point(0, N - e[m - 1]));    ends.add(new Point(e[m - 1], N));    ends.add(new Point(N, e[m - 1]));    for (int i = m - 2; i >= 0; -- i) {      List
       
         newEnds = new ArrayList 
        
         ();      if (d[i] > 0) {        int low = 0;        while (low + 1 < ends.size() && ends.get(low).y < ends.get(low + 1).y) {          ++low;        }        for (int j = 0; j <= low; ++ j) {          Point p = ends.get(j);          newEnds.add(new Point(p.x - d[i], p.y));        }        for (int j = low; j < ends.size(); ++ j) {          Point p = ends.get(j);          newEnds.add(new Point(p.x + d[i], p.y));        }        ends = newEnds;      }      newEnds = new ArrayList
         
          ();      for (int j = 0; j < ends.size(); ++ j) {        if ((j + 1 < ends.size() && ends.get(j + 1).x < 0) || (j > 0 && ends.get(j - 1).x > N))  {          continue;        }        Point p = ends.get(j);        newEnds.add(new Point(p.x, p.y + N - Math.abs(p.x - e[i])));        if (p.x < e[i] && j + 1 < ends.size() && e[i] < ends.get(j + 1).x) {          Point q = ends.get(j + 1);          newEnds.add(new Point(e[i], N + p.y + (q.y - p.y) * (e[i] - p.x) / (q.x - p.x)));        }      }      ends = newEnds;    }    int result = 0;;    for (Point end : ends) {      if (0 <= end.x && end.x <= N) {        result = Math.max(result, (int)end.y);      }    }    return result;  }  class Point {    Point(long x, long y) {      this.x = x;      this.y = y;    }    long x, y;  }}