《Cracking the Coding Interview》——第9章：递归和动态规划——题目10

2014-03-20 04:15

题目：你有n个盒子，用这n个盒子堆成一个塔，要求下面的盒子必须在长宽高上都严格大于上面的。如果你不能旋转盒子变换长宽高，这座塔最高能堆多高？

解法：首先将n个盒子按照长宽高顺序排好序，然后动态规划，我写了个O(n^2)时间复杂度的代码。

代码：

// 9.10 A stack of n boxes is form a tower. where every stack must be strictly larger than the one right above it.
// The boxes cannot be rotated.
#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;

struct Box {
    // width
    int w;
    // height
    int h;
    // depth
    int d;
    Box(int _w = 0, int _h = 0, int _d = 0): w(_w), h(_h), d(_d) {};
    
    bool operator < (const Box &other) {
        if (w != other.w) {
            return w < other.w;
        } else if (h != other.h) {
            return h < other.h;
        } else {
            return d < other.d;
        }
    };
};

int main()
{
    int n, i, j;
    Box box;
    vector<Box> v;
    vector<int> dp;
    int result;
    
    while (scanf("%d", &n) == 1 && n > 0) {
        v.resize(n);
        for (i = 0; i < n; ++i) {
            scanf("%d%d%d", &v[i].w, &v[i].h, &v[i].d);
        }
        sort(v.begin(), v.end());
        dp.resize(n);
        for (i = 0; i < n; ++i) {
            dp[i] = v[i].h;
        }
        for (i = 0; i < n; ++i) {
            for (j = 0; j < i; ++j) {
                if (v[i].w > v[j].w && v[i].h > v[j].h && v[i].d > v[j].d) {
                    dp[i] = v[i].h + dp[j] > dp[i] ? v[i].h + dp[j] : dp[i];
                }
            }
        }
        result = dp[0];
        for (i = 1; i < n; ++i) {
            result = dp[i] > result ? dp[i] : result;
        }
        v.clear();
        dp.clear();
        printf("%d
", result);
    }
    
    return 0;
}