《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：

 // 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?
 #include <cstdio>
 using namespace std;

 int combination(int n, int k)
 {
     if (n < k) {
         return ;
     } else if (n == ) {
         return ;
     } else if (k > n / ) {
         return combination(n, n - k);
     }

     int i;
     int res, div;

     res = div = ;
     for (i = ; i <= k; ++i) {
         res *= (n +  - i);
         div *= i;
         if (res % div == ) {
             res /= div;
             div = ;
         }
     }
     if (res % div == ) {
         res /= div;
         div = ;
     }

     return res;
 }

 int main()
 {
     int x, y;

     while (scanf("%d%d", &x, &y) ==  && x >=  && y >= ) {
         printf("%d\n", combination(x + y, x));
     }

     return ;
 }

解法2：如果有障碍的话，用动态规划。

代码：

 // 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?
 // If some of the positions are off limits, what would you do?
 #include <cstdio>
 #include <vector>
 using namespace std;

 int main()
 {
     int x, y;
     int i, j;
     vector<vector<int> > off_limits;
     vector<vector<int> > res;

     while (scanf("%d%d", &x, &y) ==  && x >=  && y >= ) {
         ++x;
         ++y;
         off_limits.resize(x);
         res.resize(x);
         for (i = ; i < x; ++i) {
             off_limits[i].resize(y);
             res[i].resize(y);
         }
         for (i = ; i < x; ++i) {
             for (j = ; j < y; ++j) {
                 scanf("%d", &off_limits[i][j]);
                 off_limits[i][j] = !!off_limits[i][j];
                 res[i][j] = ;
             }
         }

         res[][] = off_limits[][] ?  : ;
         for (i = ; i < x; ++i) {
             res[i][] = off_limits[i][] ?  : res[i - ][];
         }
         for (j = ; j < y; ++j) {
             res[][j] = off_limits[][j] ?  : res[][j - ];
         }
         for (i = ; i < x; ++i) {
             for (j = ; j < y; ++j) {
                 res[i][j] = off_limits[i][j] ?  : res[i - ][j] + res[i][j - ];
             }
         }

         for (i = ; i < x; ++i) {
             for (j = ; j < y; ++j) {
                 printf("%d ", res[i][j]);
             }
             printf("\n");
         }
         printf("%d\n", res[x - ][y - ]);

         for (i = ; i < x; ++i) {
             off_limits[i].clear();
             res[i].clear();
         }
         off_limits.clear();
         res.clear();
     }

     return ;
 }