HDU1069 Monkey and Banana ——dp

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16077    Accepted Submission(s): 8537

Problem Description

A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If the monkey is clever enough, it shall be able to reach the banana by placing one block on the top another to build a tower and climb up to get its favorite food.

The researchers have n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions (xi, yi, zi). A block could be reoriented so that any two of its three dimensions determined the dimensions of the base and the other dimension was the height. 

They want to make sure that the tallest tower possible by stacking blocks can reach the roof. The problem is that, in building a tower, one block could only be placed on top of another block as long as the two base dimensions of the upper block were both strictly smaller than the corresponding base dimensions of the lower block because there has to be some space for the monkey to step on. This meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked. 

Your job is to write a program that determines the height of the tallest tower the monkey can build with a given set of blocks.

 

Input

The input file will contain one or more test cases. The first line of each test case contains an integer n,
representing the number of different blocks in the following data set. The maximum value for n is 30.
Each of the next n lines contains three integers representing the values xi, yi and zi.
Input is terminated by a value of zero (0) for n.

 

Output

For each test case, print one line containing the case number (they are numbered sequentially starting from 1) and the height of the tallest possible tower in the format "Case case: maximum height = height".

 

Sample Input

110 20 3026 8 105 5 571 1 12 2 23 3 34 4 45 5 56 6 67 7 7531 41 5926 53 5897 93 2384 62 6433 83 270

 

Sample Output

Case 1: maximum height = 40Case 2: maximum height = 21Case 3: maximum height = 28Case 4: maximum height = 342

 

Source

University of Ulm Local Contest 1996

题意：有n个长方体，每个长方体都有三个尺寸代表长、宽、高，如果一个长方体A的长和宽比另一个长方体B的长和宽都要小的话，那么A可以放在B的上面。每种长方体数目不限，问可以叠放的最大高度是多少？

解析：设长方体的三个尺寸分别为x,y,z，将x，y，z按递增排序，然后可以得到三个长方体，分别是(x, y, z), (x, z, y), (y, z, x)。将得到的所有长方体进行排序，长和宽小的在前面。然后进行两层for循环即可，状态转移方程为：

h[i] = max(h[i], h[j] + a[i])其中a[j]的长和宽都小于a[i]的长和宽。

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define N 100struct Dim{int x, y, z;bool operator <(const Dim &e)const{if(e.x != x)return x < e.x;else if(e.y != y)return y < e.y;elsereturn z > e.z;}}; Dim a[N];int h[N];void Tsort(int &x, int &y, int &z){int p = min(min(x, y), z);int q = max(max(x, y), z);int r;if(x != p && x != q) r = x;else if(y != p && y != q) r = y;else r = z;x = p, y = r, z = q;}bool cmp(const Dim &p, const Dim &q){if(p.x > q.x && p.y > q.y)return true;return false;}int main(){int n, x, y, z, cas = 0;while(scanf("%d", &n) && n){int m = 1;a[0].x = a[0].y = a[0].z = 0;for(int i = 0; i < n; i++){scanf("%d%d%d", &x, &y, &z);Tsort(x, y, z);a[m].x = x, a[m].y = y, a[m++].z = z;a[m].x = x, a[m].y = z, a[m++].z = y;a[m].x = y, a[m].y = z, a[m++].z = x;}sort(a, a + m);memset(h, 0, sizeof(h));h[0] = a[0].z;int Max = h[0];for(int i = 1; i < m; i++){for(int j = 0; j < i; j++){if(cmp(a[i], a[j])){h[i] = max(h[i], h[j] + a[i].z);}}Max = max(Max, h[i]);}printf("Case %d: maximum height = %d\n", ++cas, Max);}return 0;}

另外一个类似的题目：HDU1087

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define N 1004typedef long long ll;ll a[N], b[N];int main(){int n;ll x;while(scanf("%d", &n) && n){for(int i = 1; i <= n; i++)scanf("%lld", &a[i]);ll Max = 0;memset(b, 0, sizeof(b));a[0] = 0;for(int i = 1; i <= n; i++){for(int j = 0; j < i; j++){if(a[i] > a[j]){b[i] = max(b[i], b[j] + a[i]);Max = max(Max, b[i]);}}}printf("%lld\n", Max);}return 0;}