紫书第九章-The Tower of Babylon

The Tower of Babylon

Description

Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story: The babylonians had 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 wanted to construct the tallest tower possible by stacking blocks. The problem was 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. 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 babylonians 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

这里用的是最长递增子序列思想求解。

#include
#include
#include
#include
#include
const int maxt=55;
const int maxn=31;
const int INF=0x3f3f3f3f;
using namespace std;

int d[maxn*3];
struct node{
	int x,y,z;
	 node(int x,int y,int z):x(x),y(y),z(z){//
//		this.x=x;
//		this.y=y;
//		this.z=z;
	}
//	node(){}
	bool operator<(const node &n)const{
	return (x vec;
bool cmp(node a,node b){
	return a.x*a.y在塔上面的立方体的底面积一定比下面的小
}

int LIS(int n){
	int t=0;
	for(int i=0;it) t=d[i];
		
	}
	return t;
}
int main(){
	int n,Case=0;
	while(cin>>n&&n){
		vec.clear();
		memset(d,0,sizeof(d));
		int x,y,z;
		for(int i=0;i>x>>y>>z;
				vec.push_back(node(x,y,z));
				vec.push_back(node(x,z,y));
				vec.push_back(node(z,y,x));
			}
			sort(vec.begin(),vec.end(),cmp);
			cout<<"Case "<建图求解，可以把问题转化成求DAG上的最长路问题详解DAG请点击打开链接

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 31;
struct Node{
    int x;
    int y;
    int z;
    Node(int x,int y,int z):x(x),y(y),z(z){}
    Node(){}
    bool operator < (const Node &n)const{
        return (x < n.x&& y < n.y) || (x < n.y&& y < n.x); 
    }
};
vector vec;
int n;
int d[maxn*3];
int G[maxn*3][maxn*3];
int dp(int i,int h){
    int & ans = d[i]; 
    if(ans>0)return ans;
    ans = h;
    for(int j = 0; j < n*3; j++)if(G[i][j]){
        ans = max(ans,dp(j,vec[j].z)+h);
    }
    return ans;
}
int main(){
    int cnt = 0;
    while(scanf("%d",&n)==1&&n){
        vec.clear();
        memset(G,0,sizeof(G));
        memset(d,0,sizeof(d));
        for(int i = 0; i< n; i++){
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            vec.push_back(Node(x,y,z));
            vec.push_back(Node(x,z,y));
            vec.push_back(Node(z,y,x));
        }
        sort(vec.begin(),vec.end());

        for(int i = 0;i < n*3; i++){
            for(int j = 0; j < n*3; j++){
                if(vec[i] < vec[j])
                    G[i][j] = 1;
            }
        }
        int result = -1;
        for(int i = 0; i < n*3 ;i++){
            result = max(result,dp(i,vec[i].z));
        }
        printf("Case %d: maximum height = %d\n",++cnt,result);
    }
    return 0;
}