《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 = , int _h = , int _d = ): 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) ==  && n > ) {
         v.resize(n);
         for (i = ; 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 = ; i < n; ++i) {
             dp[i] = v[i].h;
         }
         for (i = ; i < n; ++i) {
             for (j = ; 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[];
         for (i = ; i < n; ++i) {
             result = dp[i] > result ? dp[i] : result;
         }
         v.clear();
         dp.clear();
         printf("%d\n", result);
     }

     return ;
 }