《Cracking the Coding Interview》——第11章：排序和搜索——题目7

2014-03-21 22:05

题目：给你N个盒子堆成一座塔，要求下面盒子的长和宽都要严格大于上面的。问最多能堆多少个盒子？

解法1：O(n^2)的动态规划解决。其实是最长递增子序列问题，所以也可以用O(n * log(n))的优化算法。

代码：

// 11.7 n boxes are to stack up to a tower. Every box must be strictly smaller in width and height than the one right below it.

// How many boxes at most can you stack up?

#include <algorithm>

#include <cstdio>

#include <vector>

using namespace std;



struct Box {

    int x;

    int y;

    Box(int _x = 0, int _y = 0): x(_x), y(_y) {};

    

    bool operator < (const Box &other) {

        if (x != other.x) {

            return x < other.x;

        } else {

            return y < other.y;

        }

    };

};



int main()

{

    vector<Box> v;

    vector<int> dp;

    int n;

    int i, j;

    int res;

    

    while (scanf("%d", &n) == 1 && n > 0) {

        v.resize(n);

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

            scanf("%d%d", &v[i].x, &v[i].y);

        }

        sort(v.begin(), v.end());

        dp.resize(n);

        res = 0;

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

            dp[i] = 1;

            for (j = 0; j < i; ++j) {

                if (v[j].x < v[i].x && v[j].y < v[i].y) {

                    dp[i] = dp[j] + 1 > dp[i] ? dp[j] + 1 : dp[i];

                }

            }

            res = dp[i] > res ? dp[i] : res;

        }

        printf("%d\n", res);

        

        v.clear();

        dp.clear();

    }

    

    return 0;

}

解法2：用二分查找优化后的代码，其中使用了STL算法库提供的lower_bound()，二分也不总是要手写的。

代码：

 1 // 11.7 n boxes are to stack up to a tower. Every box must be strictly smaller in width and height than the one right below it.

 2 // How many boxes at most can you stack up?

 3 #include <algorithm>

 4 #include <cstdio>

 5 #include <vector>

 6 using namespace std;

 7 

 8 struct Box {

 9     int x;

10     int y;

11     Box(int _x = 0, int _y = 0): x(_x), y(_y) {};

12     

13     bool operator < (const Box &other) {

14         if (x != other.x) {

15             return x < other.x;

16         } else {

17             return y < other.y;

18         }

19     };

20 };

21 

22 int main()

23 {

24     vector<Box> v;

25     vector<int> dp;

26     vector<int>::iterator it;

27     int n;

28     int i;

29     

30     while (scanf("%d", &n) == 1 && n > 0) {

31         v.resize(n);

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

33             scanf("%d%d", &v[i].x, &v[i].y);

34         }

35         sort(v.begin(), v.end());

36         dp.push_back(v[0].y);

37         for (i = 1; i < n; ++i) {

38             if (v[i].y > dp[dp.size() - 1]) {

39                 dp.push_back(v[i].y);

40             } else {

41                 it = lower_bound(dp.begin(), dp.end(), v[i].y);

42                 *it = v[i].y;

43             }

44         }

45         printf("%d\n", (int)dp.size());

46         

47         v.clear();

48         dp.clear();

49     }

50     

51     return 0;

52 }