Painter
Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total Submission (s): 133 Accepted Submission (s):
Problem DescriptionMr. Hdu is an painter, as we all know, painters need ideas to innovate, one day, he got stuck in rut and the ideas Dr Y up, he took out a drawing board and began to draw casually. Imagine The board is a rectangle, consists of several square grids. He drew diagonally, so there is a kinds of draws, one is like ' \ ', and the other was like '/'. In each draw he choose arbitrary number of grids to draw. He always drew the first kind in red color, and drew the other kind in blue color, when a grid was drew by both Red and Blu E, it becomes green. A Grid would never is drew by the same color more than one time. Now give your The ultimate state of the board, can you calculate the minimum time of draws to reach this state. InputThe first line was an integer T describe the number of test cases.
Each test case begins with an integer number n describe the number of the rows of the drawing board.
Then n lines of string consist of ' R ' B ' G ' and '. ' of the same length. '. ' means the grid has not been drawn.
1<=n<=50
The number of column of the rectangle is also less than 50.
Output
Output an integer as described in the problem description.
OutputOutput An integer as described in the problem description. Sample Input
24RR. B.rg. Brrb. R4rrbbrggbbggrbbrr
Sample Output
36
Sourcemulti-university Training Contest 3
Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5319
The main idea: Brush the wall, can only press two oblique 45 degrees brush, one is red, one is blue, red meets blue to green, each lattice the same color only brush once, each brush must be a continuous paragraph, ask the answer target State at least a few times
Title analysis: Enumeration in two directions, \ This look like brush, if the current point is red or green and oblique before a point is not red and not green, it is necessary to brush once, the same way/so brush to judge, but why this is the smallest, because I only have to brush when the brush, so obviously this is the best
#include <cstdio> #include <cstring>char s[55][55];int main () { int T; scanf ("%d", &t); while (t--) { int n; scanf ("%d", &n); for (int i = 1; I <= n; i++) scanf ("%s", S[i] + 1); int m = strlen (s[1] + 1); int ans = 0; for (int i = 1, i <= N; i++) for (int j = 1; j <= M; j + +) if (s[i][j] = = ' R ' | | s[i][j] = = ' G ') if (! ( S[i-1][j-1] = = ' R ' | | S[i-1][j-1] = = ' G ')) ans + +; for (int i = 1, i <= N; i++) for (int j = 1; j <= M; j + +) if (s[i][j] = = ' B ' | | s[i][j] = = ' G ') if (! ( S[i-1][j + 1] = = ' B ' | | S[i-1][j + 1] = = ' G ')) ans + +; printf ("%d\n", ans);} }
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HDU 5319 Painter (simulated brain problems)
