# at t 5g coverage map

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### [ACM] zoj 3209 treasure map (precise coverage of dancing links, rectangular coverage)

need to achieve this. If it is impossible to make one complete map, just output-1. Sample Input 35 5 10 0 5 55 5 20 0 3 52 0 5 530 30 50 0 30 100 10 30 200 20 30 300 0 15 3015 0 30 30 Sample output 1-12 Hint For sample 1, the only piece is a complete map. For sample 2, the two pieces may overlap with each other, so you can not make a complete treasure map. For s

### ZOJ 3209 Treasure Map (precise DLX coverage)

ZOJ 3209 Treasure Map (precise DLX coverage) Your boss once had got extends copies of a treasure map. unfortunately, all the copies are now broken to your rectangular pieces, and what make it worse, he has lost some of the pieces. luckily, it is possible to figure out the position of each piece in the original map. now

### ZOJ 3209 Treasure Map (precise DLX coverage), zojdlx

ZOJ 3209 Treasure Map (precise DLX coverage), zojdlx Your boss once had got extends copies of a treasure map. unfortunately, all the copies are now broken to your rectangular pieces, and what make it worse, he has lost some of the pieces. luckily, it is possible to figure out the position of each piece in the original map

### Zoj 3209 treasure map (precise coverage of dancing links)

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

### POJ-2226 Muddy fields Binary map minimum point coverage

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 =

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### POJ-3041 asteroids the minimum point coverage of the binary map

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

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