hdu 1069 Monkey and Banana 记忆化搜索

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10239    Accepted Submission(s): 5319

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

 

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这个类似的问题，做过几次了，我记得第一次做的时候，只是想着从dp[]的意义出发去解决；现在做，清晰的根据大小直接关系建立树的结构，问题迎刃而解。

最熟悉树上的动归了。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI 3.1415926535897932384626#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   num<<1,le,mid#define rson    num<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     secondusing namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;const int maxn=  30+5   ;//const int maxm=    ;int n,N,cnt;struct Block{    int x,y,z;    Block(){}    Block(int x,int y,int z):x(x),y(y),z(z){}} block[3*maxn];vector<int > G[3*maxn];int dp[3*maxn];void add_edge(int a,int b ){    Block & b1=block[a];    Block & b2=block[b];    if(b1.x<b2.x&&b1.y<b2.y||b1.x<b2.y&&b1.y<b2.x)  G[b].push_back(a);    if(b1.x>b2.x&&b1.y>b2.y||b1.x>b2.y&&b1.y>b2.x)  G[a].push_back(b);}void init(){    for(int i=1;i<=N;i++)        G[i].clear();}int DP(int x){    if(dp[x]!=-1)  return dp[x];    dp[x]=block[x].z;    for(int i=0;i<G[x].size();i++)    {        int y=G[x][i];        dp[x]=max(dp[x],block[x].z+DP(y));    }    return dp[x];}int main(){    int x,y,z,kase=0;    while(~scanf("%d",&n)&&n)    {        N=3*n;cnt=0;        init();        for(int i=1;i<=n;i++)        {            scanf("%d%d%d",&x,&y,&z);            block[++cnt]=Block(x,y,z);            block[++cnt]=Block(x,z,y);            block[++cnt]=Block(y,z,x);        }        for(int i=1;i<=N;i++)        {            for(int j=1;j<=N;j++)            {                if(i==j)  continue;                add_edge(i,j);            }        }        int highest=0;        memset(dp,-1,sizeof dp);        for(int i=1;i<=N;i++)        {            highest=max(highest,DP(i));        }        printf("Case %d: maximum height = %d\n",++kase,highest);    }    return 0;}