1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 100;
 4 
 5 int block[MAXN][3];
 6 int dp[MAXN][MAXN];
 7 
 8 int main()
 9 {
10     int n, kase = 0;
11     while(scanf("%d",&n) && n)
12     {
13         int N = 0;
14         for(int i = 1; i<=n; i++)
15         {
16             int a, b, h;
17             scanf("%d%d%d", &a, &b, &h);
18             block[++N][0] = min(b,h), block[N][1] = max(b,h), block[N][2] = a;
19             block[++N][0] = min(a,h), block[N][1] = max(a,h), block[N][2] = b;
20             block[++N][0] = min(a,b), block[N][1] = max(a,b), block[N][2] = h;
21         }
22 
23         int ans = -2000000;
24         memset(dp, 0, sizeof(dp));
25         for(int i = 1; i<=N; i++)   //初始化第一个
26             dp[1][i] = block[i][2], ans = max(dp[1][i], ans);
27 
28         for(int i = 2; i<=N; i++)   //第i个
29         for(int j = 1; j<=N; j++)   //第i个放编号为j的块
30         for(int k = 1; k<=N; k++)   //j能否放在k上
31              if(block[j][0]<block[k][0] && block[j][1]<block[k][1])
32                 dp[i][j] = max(dp[i][j], dp[i-1][k]+block[j][2]), ans = max(dp[i][j], ans);
33 
34         printf("Case %d: maximum height = %d\n", ++kase, ans);
35     }
36     return 0;
37  }

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 100;
 4 
 5 struct node
 6 {
 7     int a, b, h;
 8     bool operator<(const node x){
 9         return a>x.a;
10     }
11 }block[MAXN];
12 int dp[MAXN];
13 
14 int main()
15 {
16     int n, kase = 0;
17     while(scanf("%d",&n) && n)
18     {
19         int N = 0;
20         for(int i = 1; i<=n; i++)
21         {
22             int a, b, h;
23             scanf("%d%d%d", &a, &b, &h);
24             block[++N].a = min(b,h), block[N].b = max(b,h), block[N].h = a;
25             block[++N].a = min(a,h), block[N].b = max(a,h), block[N].h = b;
26             block[++N].a = min(a,b), block[N].b = max(a,b), block[N].h = h;
27         }
28         sort(block+1, block+1+N);
29         int ans = -2000000;
30         for(int i = 1; i<=N; i++)    //初始化第一个
31             dp[i] = block[i].h, ans = max(ans, dp[i]);
32 
33         for(int i = 1; i<=N; i++)
34             for(int j = 1; j<i; j++)
35                 if(block[i].a<block[j].a && block[i].b<block[j].b)
36                     dp[i] = max(dp[i], dp[j]+block[i].h), ans = max(ans, dp[i]);
37 
38         printf("Case %d: maximum height = %d\n", ++kase, ans);
39     }
40     return 0;
41  }