The Tower of Babylon UVA - 437

题目链接:点我

---

    Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this talehave been forgotten. So now, in line with the educational nature of this contest, we will tell you thewhole 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(x i; 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

---

题意:

求用所给得砖块所能构成的满足方块的两个底边严格递减的条件下所能形成的最高的tower.

思路:

经典的动态规划问题,类似与最长上升子序列的做法.以下给出两种做法.

代码一:

#include
#include
#include
#include
#include
using namespace std;

const int maxn = 100;
struct ss{
    int x, y, z;
    bool operator <(const ss &q){
        return xint x, int y,int z)
    :x(x),y(y),z(z){}
}a[maxn];
int dp[maxn];

int main(){
    int n;
    int kase = 0;
    while(scanf("%d", &n) != EOF && n){
        int k = 1;
        for(int  i = 1; i <= n; ++i){
            int x, y, z;
            scanf("%d %d %d", &x, &y, &z);
            a[++k] = ss(max(x, y), min(x, y), z);//这里约定大小,可以简化过程
            a[++k] = ss(max(y, z), min(z, y), x);
            a[++k] = ss(max(z, x), min(x, z), y);
        }sort(a+1,a+1+k);
        memset(dp,0, sizeof(dp));
        int ans = 0;
        for(int  i = 1; i <= k; ++i){
            int maxi = 0;
            for(int j = i - 1; j > 0; --j)
                if(a[i].x > a[j].x && a[i].y > a[j].y && dp[j] > maxi)
                    maxi = dp[j];
            dp[i] = maxi + a[i].z;
            ans = max(ans, dp[i]);
        }
        printf("Case %d: maximum height = %d\n", ++kase,ans);
    }return 0;
}

代码二:

#include
#include
#include
#include
#include
using namespace std;

const int maxn = 1000;
struct ss{
    int x, y, z;
    bool operator <(const ss q){
        return x  < q.x;
    }ss (){};
    ss(int x, int y,int z)
    :x(x),y(y),z(z){}
}a[maxn];
int dp[maxn];

int main(){
    int n;
    int kase = 0;
    while(scanf("%d", &n) != EOF && n){
        int k = 1;
        for(int  i = 1; i <= n; ++i){
            int x, y, z;
            scanf("%d %d %d", &x, &y, &z);
            a[++k] = ss(x, y, z);//直接枚举六种不同的状态
            a[++k] = ss(y, z, x);
            a[++k] = ss(z, x, y);
            a[++k] = ss(y, x, z);
            a[++k] = ss(z, y, x);
            a[++k] = ss(x, z, y);
        }sort(a+1,a+1+k);
        memset(dp,0, sizeof(dp));
        int maxx = 0;
        for(int  i = 1; i <= k; ++i){
            dp[i] = a[i].z;
            for(int j = 1; j < i; ++j)
            if(a[i].x >a[j].x && a[i].y > a[j].y)
                dp[i] = max(dp[j] + a[i].z,dp[i]);
            maxx = max(maxx,dp[i]);
        }printf("Case %d: maximum height = %d\n", ++kase,maxx);
    }return 0;
}