《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 = , int _y = ): 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) ==  && n > ) {
        v.resize(n);
        for (i = ; i < n; ++i) {
            scanf("%d%d", &v[i].x, &v[i].y);
        }
        sort(v.begin(), v.end());
        dp.resize(n);
        res = ;
        for (i = ; i < n; ++i) {
            dp[i] = ;
            for (j = ; j < i; ++j) {
                if (v[j].x < v[i].x && v[j].y < v[i].y) {
                    dp[i] = dp[j] +  > dp[i] ? dp[j] +  : dp[i];
                }
            }
            res = dp[i] > res ? dp[i] : res;
        }
        printf("%d\n", res);

        v.clear();
        dp.clear();
    }

    return ;
}

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：

 // 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 = , int _y = ): 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;
     vector<int>::iterator it;
     int n;
     int i;

     while (scanf("%d", &n) ==  && n > ) {
         v.resize(n);
         for (i = ; i < n; ++i) {
             scanf("%d%d", &v[i].x, &v[i].y);
         }
         sort(v.begin(), v.end());
         dp.push_back(v[].y);
         for (i = ; i < n; ++i) {
             if (v[i].y > dp[dp.size() - ]) {
                 dp.push_back(v[i].y);
             } else {
                 it = lower_bound(dp.begin(), dp.end(), v[i].y);
                 *it = v[i].y;
             }
         }
         printf("%d\n", (int)dp.size());

         v.clear();
         dp.clear();
     }

     return ;
 }