题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1069

分析：每一个方块可以分成6种摆放情况，对于每种情况只能使用一次（严格小于）；方块h作为权值不需要看，从大到小排长（排宽一样）

 1 #include<iostream>
 2 #include<sstream>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<string>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<functional>
 9 #include<iomanip>
10 #include<numeric>
11 #include<cmath>
12 #include<queue>
13 #include<vector>
14 #include<set>
15 #include<cctype>
16 #define PI acos(-1.0)
17 const int INF = 0x3f3f3f3f;
18 const int NINF = -INF - 1;
19 const int maxn = 1e6 + 5;
20 typedef long long ll;
21 using namespace std;
22 int n;
23 struct node
24 {
25     int x, y, h;
26 }cube[200];
27 bool cmp(node a, node b)
28 {
29     return a.x > b.x;
30 }
31 int dp[200];
32 int main()
33 {
34     int t = 0;
35     while (scanf("%d", &n) && n)
36     {
37         t++;
38         int num = 0;
39         while (n--)
40         {
41             int a, b, c;
42             scanf("%d%d%d", &a, &b, &c);
43             cube[num].x = a, cube[num].y = b, cube[num++].h = c;
44             cube[num].x = a, cube[num].y = c, cube[num++].h = b;
45             cube[num].x = b, cube[num].y = a, cube[num++].h = c;
46             cube[num].x = b, cube[num].y = c, cube[num++].h = a;
47             cube[num].x = c, cube[num].y = b, cube[num++].h = a;
48             cube[num].x = c, cube[num].y = a, cube[num++].h = b;
49         }
50         sort(cube, cube + num, cmp);
51         memset(dp, 0, sizeof(dp));
52         dp[0] = cube[0].h;
53         for (int i = 1; i < num; ++i)
54         {
55             dp[i] = cube[i].h;
56             for (int j = i - 1; j >= 0; j--)
57             {
58                 if (cube[j].x > cube[i].x && cube[j].y > cube[i].y)
59                     dp[i] = max(dp[i], dp[j] + cube[i].h);
60             }
61         }
62         //for (int i = 0; i < num; ++i)
63         //    cout << dp[i] << ' ';
64         //cout << endl;
65         int ans = 0;
66         for (int i = 0; i < num; ++i)
67             ans = max(ans, dp[i]);
68         //cout << "Case " << t << ": maximum height = " << ans << endl;
69         printf("Case %d: maximum height = %d\n", t, ans);
70     }
71     return 0;
72 }

转载于:https://www.cnblogs.com/veasky/p/11294960.html

原文链接：https://blog.csdn.net/weixin_30342827/article/details/99083077

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