hdu 1069 动规 Monkey and Banana

 Monkey and Banana

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d
& %I64u

Submit Status Practice HDU
1069

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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct s
{
    int l,w,h;
} a[111];
int dp[111];
int cmp(s A,s B)
{
    if(A.l==B.l)
        return A.w>B.w;
    return A.l>B.l;
}
int main()
{
    int d[3],n,i,j,cot=1,k,sum;
    while(scanf("%d",&n)!=EOF&&n)
    {
        k=0;
        for(i=0; i<n; i++)
        {
            scanf("%d%d%d",&d[0],&d[1],&d[2]);
            sort(d,d+3);
            //将数据转换成多种形式的矩形体
            a[k].l=d[2];
            a[k].w=d[1];
            a[k].h=d[0];
            k++;
            a[k].l=d[2];
            a[k].w=d[0];
            a[k].h=d[1];
            k++;
            a[k].l=d[1];
            a[k].w=d[0];
            a[k].h=d[2];
            k++;
        }
        sort(a,a+k,cmp);
        for(i=0; i<k; i++) dp[i]=a[i].h;
        for(i=k-2; i>=0; i--)
            for(j=i+1; j<k; j++)
            {
                if(a[i].l>a[j].l&&a[i].w>a[j].w)//最大递减dp
                    if(dp[i]<dp[j]+a[i].h)
                        dp[i]=dp[j]+a[i].h;
            }
        sum=dp[0];
        for(i=0; i<k; i++)
            if(sum<dp[i]) sum=dp[i];
        printf("Case %d: maximum height = %d\n",cot++,sum);
    }
    return 0;
}

矩形嵌套

时间限制：3000 ms  |  内存限制：65535 KB

难度：4

描述

有n个矩形，每个矩形可以用a,b来描述，表示长和宽。矩形X(a,b)可以嵌套在矩形Y(c,d)中当且仅当a<c,b<d或者b<c,a<d（相当于旋转X90度）。例如（1,5）可以嵌套在（6,2）内，但不能嵌套在（3,4）中。你的任务是选出尽可能多的矩形排成一行，使得除最后一个外，每一个矩形都可以嵌套在下一个矩形内。

输入

第一行是一个正正数N(0<N<10)，表示测试数据组数，

每组测试数据的第一行是一个正正数n，表示该组测试数据中含有矩形的个数(n<=1000)

随后的n行，每行有两个数a,b(0<a,b<100)，表示矩形的长和宽

输出

每组测试数据都输出一个数，表示最多符合条件的矩形数目，每组输出占一行

样例输入

1
10
1 2
2 4
5 8
6 10
7 9
3 1
5 8
12 10
9 7
2 2

样例输出

5

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;

int t,n;

struct node
{
    int l,w;
} a[1001];
int dp[1001];
int cmp(node A,node B)
{
    if(A.l==B.l)
        return A.w<B.w;
    return A.l<B.l;
}

int main()
{
    int d[3];
    scanf("%d",&t);
    while(t--)
    {
        int i,j,k=0,ct=0;
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        for(i=0; i<n; i++)
        {
            scanf("%d %d",&d[0],&d[1]);
            sort(d,d+2);
            a[k].l=d[1];
            a[k].w=d[0];
            k++;
            a[k].l=d[0];
            a[k].w=d[1];
            k++;
        }
        sort(a,a+k,cmp);
        for(i=0; i<k; i++)
        {
            dp[i]=1;
            for(j=0; j<i; j++)
            {
                if(a[i].l>a[j].l&&a[i].w>a[j].w)
                {
                    dp[i]=max(dp[i],dp[j]+1);
                }
            }
        }
        int maxx=dp[0];
        for(i=0; i<k; i++)
        {
            if(dp[i]>maxx)
                maxx=dp[i];
        }
        printf("%d\n",maxx);
    }
    return 0;
}