boingo coverage map

A rectangle of N * m is given. The coordinates are (0, 0 )~ (N, m ). Then we will give you a small P rectangle with the coordinates (x1, Y1 )~ (X2, Y2), you can select the smallest rectangle so that these rectangles can overwrite the entire rectangle without overlap. (N, m Idea: precise coverage of dancing links. We divide the N * m rectangle into N * m small squares, so we only need to ensure that each small square is overwritten and only once. The

columns#include #include #include using namespace STD;Const intN =1010;Const intMAXN = -; vectorint>Row[n];intN, M, Num_r, NEW_ROW[MAXN][MAXN], NEW_COL[MAXN][MAXN], vis[n], link[n];CharSTR[MAXN][MAXN];voidInit () {Num_r =1; for(inti =1; I scanf('%s ', Str[i]);BOOLFlag for(inti =1; I if(!flag) num_r++; Flag =true; for(intj =0; J if(Str[i][j] = =' * ') {new_row[i][j+1] = Num_r; Flag =false; }Else if(Str[i][j] = ='. '!flag) {num_r++; Flag =true; } } } for(inti =1; I intCNT =1; for(inti =

The main idea: in an n * n grid, there are m obstacles, and now you have a weapon, this weapon can eliminate any line or a row of obstacles, now requires all obstacles to eliminate, ask at least how many times the use of this weaponProblem-solving ideas: Think of long time how to solve the problems of row and column ... How to represent two point setsFinally suddenly thought, since do not know how to deal with the ranks, the ranks are divided into two points, the point on behalf of the relations