SRM 631 DIV1

SRM 631 DIV1
SRM 631 DIV1

A: A maximum of two steps are required. The two rows in the middle are black and one line is white. In this case, you only need to consider the first step, and enumerate a row to dye the satisfied situation, just brute force.

B: greedy. A record records the current location of a cat and the current location of more than one cat, and then sorts the positions from left to right. If the current location can be moved to more than two positions, move all the data in the past without increasing the number. If not, consider whether the current data can be paved. If yes, update the location accordingly. If not, heap all the cats to the right, then the number of heaps is 1

Code:

A:

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     using namespace std;class TaroJiroGrid {public:bool judge(vector
      
        grid) {for (int i = 0; i < grid.size(); i++) {int cnt = 1;for (int j = 1; j < grid.size(); j++) {if (grid[j][i] == grid[j - 1][i]) {cnt++;} else {if (cnt > grid.size() / 2) return false;cnt = 1;}}if (cnt > grid.size() / 2) return false;}return true;}bool solve(vector
       
         grid, int cnt) {if (cnt == 0)if (judge(grid)) return true;else if (cnt == 1) {for (int i = 0; i < grid.size(); i++) {vector
        
          tmp = grid;for (int j = 0; j < grid[i].length(); j++)tmp[i][j] = 'B';if (judge(tmp)) return true;tmp = grid;for (int j = 0; j < grid[i].length(); j++)tmp[i][j] = 'W';if (judge(tmp)) return true;}}return false;}int getNumber(vector
         
           grid) {for (int i = 0; i < 2; i++) {if (solve(grid, i))return i;}return 2;}};
         
        
       
      
    
   
  
 

B:

#include 
 
  #include using namespace std;typedef pair
  
    pii;#define MP(a,b) make_pair(a,b)const int INF = 0x3f3f3f3f;class CatsOnTheLineDiv1 {vector
   
     g;public:int getNumber(vector
    
      position, vector
     
       count, int time) {int n = position.size();for (int i = 0; i < n; i++)g.push_back(MP(position[i] - time, count[i]));sort(g.begin(), g.end());int le = -INF, sink = -INF, ans = 0;for (int i = 0; i < n; i++) {int l = g[i].first;int r = l + 2 * time;if (l <= sink) continue;le = max(le, l);if (r - l + 1 < count[i]) {ans++;sink = r;} else {le += count[i];}}return ans;}};