《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：

// 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;
    vector<int>::iterator it;
    int n;
    int i;
    
    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.push_back(v[0].y);
        for (i = 1; i < n; ++i) {
            if (v[i].y > dp[dp.size() - 1]) {
                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 0;
}