经典dp
乍看好像是完全背包 因为无限个
创新互联公司是专业的西峡网站建设公司，西峡接单;提供成都网站制作、成都做网站,网页设计,网站设计,建网站,PHP网站建设等专业做网站服务;采用PHP框架,可快速的进行西峡网站开发网页制作和功能扩展;专业做搜索引擎喜爱的网站,专业的做网站团队,希望更多企业前来合作!
但是其实每个砖头只有三种情况 将这些情况都当成一个新的砖头放进brick里面就可以了
这里要注意 一开始我的判断是下面砖头的x y 分别大于 上面的 x y 这样可以 但要在加入exchange 中做点修改
使得每个brick 的 x 都小于 y
例如 brickA 3 9 12 brickB 4 2 13
则 brickA 会加入 9 12 3， 12 3 9 （旋转加入） brickB 2 13 4， 13 4 2
如果不限制 x < y 将检测不到 3 9 12和 2 4 13这种合法情况
其实就是 你check 判断的是 x y分别是否小于 但没判断 x1 < y2 y1 < x2 的情况（交叉的情况）
故每次确定将x < y 放入队列 就排除了这些不确定
View Code
Monkey and Banana
Time Limit:2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s):4895 Accepted Submission(s): 2503
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 monkeyis 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 problemis 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 jobis 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 testcase contains an integer n,
representing the number of different blocksin 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.
Inputis terminated by a value of zero (0) for n.
Output
For each testcase, 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
Case1: maximum height = 40
Case2: maximum height = 21
Case3: maximum height = 28
Case4: maximum height = 342
View Code
#include
#include
#include
using std::stack;
using std::sort;
stack s;
int cnt, dp[100], n;
struct node
{
int x, y, h;
bool operator > (node a) const
{
if( x == a.x )
return y >= a.y;
else return x > a.x ;
}
}brick[100];
void exchange(int i)
{
int x = brick[i].x;
int y = brick[i].y;
int h = brick[i].h;
if( x > y )
{
int tmp = brick[i].x;
brick[i].x= brick[i].y;
brick[i].y= tmp;
}
brick[++i].x = y;
brick[i].y= h;
brick[i].h= x;
if( y > h )
{
int tmp = brick[i].x;
brick[i].x= brick[i].y;
brick[i].y= tmp;
}
brick[++i].x = h;
brick[i].y= x;
brick[i].h= y;
if( h > x )
{
int tmp = brick[i].x;
brick[i].x= brick[i].y;
brick[i].y= tmp;
}
}
int cmp(node a, node b)
{return a > b; }
bool check(int i, int j)
{
return brick[i].x < brick[j].x && brick[i].y < brick[j].y;
}
int max(int a, int b)
{return a>=b ?a :b ; }
int main()
{
int i, j, cas = 0;
while(scanf("%d", &n), n)
{
int maxNum = 0;
for(i=1; i<=n*3; i+=3)
{
scanf("%d %d %d", &brick[i].x, &brick[i].y, &brick[i].h);
exchange(i);
}
cnt= n*3;
sort(brick+1, brick+1+cnt, cmp);
for(i=1; i<=cnt; i++)
{
dp[i]= brick[i].h ;
for(j=1; j maxNum )
maxNum= dp[i];
printf("Case %d: maximum height = %d
", ++cas, maxNum);
}
return 0;
}
网页题目：MonkeyandBanana背包dp-创新互联
当前路径：
http://cdkjz.cn/article/cdhdph.html
