《Cracking the Coding Interview》——第9章：递归和动态规划——题目7

2014-03-20 03:35

题目：实现画图的Flood Fill操作。

解法：DFS和BFS皆可，但BFS使用的队列在时间复杂度上常数项比较大，速度略慢，所以我选了DFS。当然，如果图很大的话DFS是会导致call stack溢出的，那就摊上事儿了。

代码：

 1 // 9.7 Implement a flood fill painter that changes a certain area to a certain color. You are given one point as the seed.

 2 #include <cstdio>

 3 #include <vector>

 4 using namespace std;

 5 

 6 void floodFill(int i, int j, int n, int m, int new_color, vector<vector<int> > &canva)

 7 {

 8     int old_color = canva[i][j];

 9 

10     canva[i][j] = new_color;

11     if (i >= 1 && old_color == canva[i - 1][j]) {

12         floodFill(i - 1, j, n, m, new_color, canva);

13     }

14     if (i <= n - 2 && old_color == canva[i + 1][j]) {

15         floodFill(i + 1, j, n, m, new_color, canva);

16     }

17     if (j >= 1 && old_color == canva[i][j - 1]) {

18         floodFill(i, j - 1, n, m, new_color, canva);

19     }

20     if (j <= m - 2 && old_color == canva[i][j + 1]) {

21         floodFill(i, j + 1, n, m, new_color, canva);

22     }

23 }

24 

25 int main()

26 {

27     int i, j, c;

28     int n, m;

29     vector<vector<int> > canva;

30     

31     scanf("%d%d", &n, &m);

32     canva.resize(n);

33     for (i = 0; i < n; ++i) {

34         canva[i].resize(m);

35     }

36     

37     for (i = 0; i < n; ++i) {

38         for (j = 0; j < m; ++j) {

39             scanf("%d", &canva[i][j]);

40         }

41     }

42     

43     while (scanf("%d%d%d", &i, &j, &c) == 3) {

44         if (i >= 0 && i <= n - 1 && j >= 0 && j <= m - 1) {

45             floodFill(i, j, n, m, c, canva);

46         }

47         for (i = 0; i < n; ++i) {

48             for (j = 0; j < m; ++j) {

49                 printf((j == 0 ? "%d" : " %d"), canva[i][j]);

50             }

51             printf("\n");

52         }

53     }

54     

55     for (i = 0; i < n; ++i) {

56         canva[i].clear();

57     }

58     canva.clear();

59     

60     return 0;    

61 }