《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 0;
    } else if (n == 0) {
        return 0;
    } else if (k > n / 2) {
        return combination(n, n - k);
    }

    int i;
    int res, div;
    
    res = div = 1;
    for (i = 1; i <= k; ++i) {
        res *= (n + 1 - i);
        div *= i;
        if (res % div == 0) {
            res /= div;
            div = 1;
        }
    }
    if (res % div == 0) {
        res /= div;
        div = 1;
    }
    
    return res;
}

int main()
{
    int x, y;
    
    while (scanf("%d%d", &x, &y) == 2 && x >= 0 && y >= 0) {
        printf("%d
", combination(x + y, x));
    }
    
    return 0;
}

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) == 2 && x >= 0 && y >= 0) {
        ++x;
        ++y;
        off_limits.resize(x);
        res.resize(x);
        for (i = 0; i < x; ++i) {
            off_limits[i].resize(y);
            res[i].resize(y);
        }
        for (i = 0; i < x; ++i) {
            for (j = 0; j < y; ++j) {
                scanf("%d", &off_limits[i][j]);
                off_limits[i][j] = !!off_limits[i][j];
                res[i][j] = 0;
            }
        }
        
        res[0][0] = off_limits[0][0] ? 0 : 1;
        for (i = 1; i < x; ++i) {
            res[i][0] = off_limits[i][0] ? 0 : res[i - 1][0];
        }
        for (j = 1; j < y; ++j) {
            res[0][j] = off_limits[0][j] ? 0 : res[0][j - 1];
        }
        for (i = 1; i < x; ++i) {
            for (j = 1; j < y; ++j) {
                res[i][j] = off_limits[i][j] ? 0 : res[i - 1][j] + res[i][j - 1];
            }
        }
        
        for (i = 0; i < x; ++i) {
            for (j = 0; j < y; ++j) {
                printf("%d ", res[i][j]);
            }
            printf("
");
        }
        printf("%d
", res[x - 1][y - 1]);
        
        for (i = 0; i < x; ++i) {
            off_limits[i].clear();
            res[i].clear();
        }
        off_limits.clear();
        res.clear();
    }
    
    return 0;
}