(DP,最长上升子序列变形)Monkey and Banana--HDOJ
来源:互联网 发布:php 图片转base64 编辑:程序博客网 时间:2024/06/06 02:52
Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1059 Accepted Submission(s): 643
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
1
10 20 30
2
6 8 10
5 5 5
7
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
5
31 41 59
26 53 58
97 93 23
84 62 64
33 83 27
0
Sample Output
Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
Source
University of Ulm Local Contest 1996
Recommend
JGShining
总结:
思路很快就出来啦,我也很自信我能1A,可是,并不是我想象的那样,
我的mx没赋初值,W了好几发。。。。
#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>using namespace std;typedef struct Node{ int xx,yy,hh;}Node;bool cmp(Node A,Node B){ if(A.xx != A.yy) return A.xx < B.xx; else return A.yy < B.yy;}int main(){ // freopen("in.txt","r",stdin); int n,st=1; while(scanf("%d",&n) && n) { int x,y,z,cnt=0; Node num[200]; for(int i=0; i<n; ++i) { scanf("%d %d %d",&x,&y,&z); num[cnt].xx = x,num[cnt].yy = y,num[cnt++].hh = z; num[cnt].xx = y,num[cnt].yy = x,num[cnt++].hh = z; num[cnt].xx = x,num[cnt].yy = z,num[cnt++].hh = y; num[cnt].xx = z,num[cnt].yy = x,num[cnt++].hh = y; num[cnt].xx = y,num[cnt].yy = z,num[cnt++].hh = x; num[cnt].xx = z,num[cnt].yy = y,num[cnt++].hh = x; } sort(num,num+cnt,cmp); int dp[200],mx=-1; memset(dp,0,sizeof(dp)); for(int i=0; i<cnt; ++i) { dp[i] = num[i].hh; for(int j=0; j<i; ++j) { if(num[i].xx > num[j].xx && num[i].yy > num[j].yy) { if(dp[i] < dp[j] + num[i].hh) { dp[i] = dp[j] + num[i].hh; mx = max(dp[i],mx); } } } mx = max(dp[i],mx); } printf("Case %d: maximum height = %d\n",st,mx); st++; } return 0;}
- (DP,最长上升子序列变形)Monkey and Banana--HDOJ
- hdu 1069 Monkey and Banana(dp 最长上升子序列)
- HDU 1069 Monkey and Banana (dp, 最长上升子序列)
- hdu 1069 Monkey and Banana(类似最长上升子序列,dp)
- HDU 1069 Monkey and Banana(dp最长上升子序列)
- HDU 1069 Monkey and Banana (最长上升子序列)
- hdu 1069 Monkey and Banana 最长上升子序列。
- HDU 1069 Monkey and Banana 最长上升子序列模板
- HDU 1069 Monkey and Banana dp类型:最长上升子序列
- Monkey and Banana (最长子序列)
- HDU1069 Monkey and Banana(dp动态规划,最长非递减子序列变形题)
- 动态规划1:H - Monkey and Banana(最长非上升子序列)
- hdoj 1069 Monkey and Banana(上升子序列最大和)
- hdu 1069 monkey and banana(最长下降子序列)
- hdu 1069 Monkey and Banana(最长递减子序列 )
- 三维最长上升子序列问题——HDU 1069 Monkey and Banana
- hdu 1069 Monkey and Banana(最长递增子序列的变形)
- hdu 1069 Monkey and Banana 再来一波DP啦~~LIS的变形,会最大上升子序列就可以A啦~
- 第14节-Linux支持的其他文件系统与 VFS
- 浅谈c++volatile关键字
- 管理感言_打铁还需自身硬
- 分享AndroidStudio相关的一些安装和使用的链接
- 元素出栈入栈的合法性
- (DP,最长上升子序列变形)Monkey and Banana--HDOJ
- 接口测试-py-post
- selenium 错误集锦->f.QueryInterface is not a function
- vs2013中定义的全局变量count在使用时提示:“不明确的符号”
- Linux学习笔记--初识
- 用 线性回归 预测股票的涨跌
- [IIOT学习]Microzed IIOT开发板连接IBM bluemix云
- String常用的构造方法
- java对象赋值