UVa 437 The Tower of Babylon

和之前的 Stacking Boxes 很像,但是这里求的是最大高度,并且盒子成了立体状,因此对每个盒子有三种不同的高度,分别为x, y, z,相当于增加两个盒子;

记忆化搜索。

 1 # include <stdio.h>
2 # include <memory.h>
3
4 typedef struct {
5 int x;
6 int y;
7 int h;
8 }block;
9
10 int n;
11 int f[91];
12 block b[91];
13
14 int cmp(const void *x, const void *y)
15 {
16 return (*(int *)x > *(int *)y ? 1:-1);
17 }
18
19 int dp(int i);
20
21 int main()
22 {
23 int i, cnt, maxH, tmp, t[3];
24
25 cnt = 0;
26 while (1)
27 {
28 ++cnt;
29 scanf("%d", &n);
30 if ( !n ) break;
31 n *= 3;
32 for ( i = 1; i <= n; ++i)
33 {
34 scanf("%d%d%d", &t[0], &t[1], &t[2]);
35 qsort(t, 3, sizeof(int), cmp);
36 b[i].x = t[0]; b[i].y = t[1]; b[i].h = t[2]; ++i;
37 b[i].x = t[0]; b[i].y = t[2]; b[i].h = t[1]; ++i;
38 b[i].x = t[1]; b[i].y = t[2]; b[i].h = t[0];
39 }
40
41 maxH = 0;
42 memset(f, 0, sizeof(f));
43 for ( i = 1; i <= n; ++i)
44 if (maxH < (tmp = dp(i))) maxH = tmp;
45
46 printf("Case %d: maximum height = %d\n", cnt, maxH);
47 }
48
49 return 0;
50 }
51
52 int dp(int i)
53 {
54 int j, tmp;
55
56 if (f[i] > 0) return f[i];
57 f[i] = b[i].h;
58 for ( j = 1; j <= n; ++j)
59 {
60 if (b[i].x>b[j].x && b[i].y>b[j].y && f[i] < (tmp = dp(j)+b[i].h))
61 f[i] = tmp;
62 }
63
64 return f[i];
65 }
原文地址:https://www.cnblogs.com/JMDWQ/p/2431967.html