《Cracking the Coding Interview》——第9章：递归和动态规划——题目2

2014-03-20 02:55

题目：从(0, 0)走到(x, y)，其中x、y都是非负整数。每次只能向x或y轴的正方向走一格，那么总共有多少种走法。如果有些地方被障碍挡住不能走呢？

解法1：如果没有障碍的话，组合数学，排列组合公式C(x + y, x)。

代码：

 1 // 9.2 How many ways are there to go from (0, 0) to (x, y), if you only go left or up, and one unit at a time?

 2 #include <cstdio>

 3 using namespace std;

 4 

 5 int combination(int n, int k)

 6 {

 7     if (n < k) {

 8         return 0;

 9     } else if (n == 0) {

10         return 0;

11     } else if (k > n / 2) {

12         return combination(n, n - k);

13     }

14 

15     int i;

16     int res, div;

17     

18     res = div = 1;

19     for (i = 1; i <= k; ++i) {

20         res *= (n + 1 - i);

21         div *= i;

22         if (res % div == 0) {

23             res /= div;

24             div = 1;

25         }

26     }

27     if (res % div == 0) {

28         res /= div;

29         div = 1;

30     }

31     

32     return res;

33 }

34 

35 int main()

36 {

37     int x, y;

38     

39     while (scanf("%d%d", &x, &y) == 2 && x >= 0 && y >= 0) {

40         printf("%d\n", combination(x + y, x));

41     }

42     

43     return 0;

44 }

解法2：如果有障碍的话，用动态规划。

代码：

 1 // 9.2 How many ways are there to go from (0, 0) to (x, y), if you only go left or up, and one unit at a time?

 2 // If some of the positions are off limits, what would you do?

 3 #include <cstdio>

 4 #include <vector>

 5 using namespace std;

 6 

 7 int main()

 8 {

 9     int x, y;

10     int i, j;

11     vector<vector<int> > off_limits;

12     vector<vector<int> > res;

13     

14     while (scanf("%d%d", &x, &y) == 2 && x >= 0 && y >= 0) {

15         ++x;

16         ++y;

17         off_limits.resize(x);

18         res.resize(x);

19         for (i = 0; i < x; ++i) {

20             off_limits[i].resize(y);

21             res[i].resize(y);

22         }

23         for (i = 0; i < x; ++i) {

24             for (j = 0; j < y; ++j) {

25                 scanf("%d", &off_limits[i][j]);

26                 off_limits[i][j] = !!off_limits[i][j];

27                 res[i][j] = 0;

28             }

29         }

30         

31         res[0][0] = off_limits[0][0] ? 0 : 1;

32         for (i = 1; i < x; ++i) {

33             res[i][0] = off_limits[i][0] ? 0 : res[i - 1][0];

34         }

35         for (j = 1; j < y; ++j) {

36             res[0][j] = off_limits[0][j] ? 0 : res[0][j - 1];

37         }

38         for (i = 1; i < x; ++i) {

39             for (j = 1; j < y; ++j) {

40                 res[i][j] = off_limits[i][j] ? 0 : res[i - 1][j] + res[i][j - 1];

41             }

42         }

43         

44         for (i = 0; i < x; ++i) {

45             for (j = 0; j < y; ++j) {

46                 printf("%d ", res[i][j]);

47             }

48             printf("\n");

49         }

50         printf("%d\n", res[x - 1][y - 1]);

51         

52         for (i = 0; i < x; ++i) {

53             off_limits[i].clear();

54             res[i].clear();

55         }

56         off_limits.clear();

57         res.clear();

58     }

59     

60     return 0;

61 }