The Tower of Babylon UVA - 437 DAG上的动态规划

Posted 2021-01-19 fan-jiaming

题目：题目链接

思路：每个方块可以用任意多次，但因为底面限制，每个方块每个放置方式选一个就够了，以x y为底 z 为高，以x z为底 y 为高，以y z为底 x为高，因为数据量很小，完全可以把每一种当成DAG上的一个结点，然后建图找最长路径。

AC代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <algorithm>
 5 #include <cstring>
 6 #include <string>
 7 #include <vector>
 8 #include <map>
 9 #include <set>
10 #include <queue>
11 #include <deque>
12 #include <stack>
13 #include <list>
14 
15 #define FRER() freopen("in.txt", "r", stdin)
16 #define FREW() freopen("out.txt", "w", stdout)
17 
18 #define INF 0x3f3f3f3f
19 
20 using namespace std;
21 
22 struct block {
23     int x, y, z;
24     block() {}
25     block(int x, int y, int z):x(x), y(y), z(z) {}
26     bool operator < (const block & other) const {
27         return (x < other.x && y < other.y) || (x < other.y && y < other.x);
28     }
29 };
30 
31 vector<block> vec;
32 
33 int n, m, d[100];
34 bool G[100][100];
35 
36 void init() {
37     vec.clear();
38     int x, y, z;
39     for(int i = 0; i < n; ++i) {
40         cin >> x >> y >> z;
41         vec.push_back(block(x, y, z));
42         vec.push_back(block(x, z, y));
43         vec.push_back(block(y, z, x));
44     }
45     m = n * 3;
46     memset(G, 0, sizeof(G));
47     for(int i = 0; i < m; ++i) 
48         for(int j = 0; j < m; ++j)
49             if(vec[i] < vec[j])
50                 G[i][j] = 1;
51 
52     memset(d, 0, sizeof(d));
53 }
54 
55 int dp(int i, int h) {
56     if(d[i]) return d[i];
57     int& ans = d[i];
58     ans = h;
59     m = n * 3;
60     for(int j = 0; j < m; ++j) {
61         if(G[i][j])
62             ans = max(ans, dp(j, vec[j].z) + h);
63     }
64     return ans;
65 }
66 
67 
68 int solve() {
69     int ans = -1;
70     for(int i = 0; i < m; ++i) 
71         ans = max(ans, dp(i, vec[i].z));
72     return ans;
73 }
74 
75 int main()
76 {
77     //FRER();
78     //FREW();
79     ios::sync_with_stdio(0);
80     cin.tie(0);
81     
82     int kase = 0;
83     while(cin >> n, n) {
84         init();
85         cout << "Case " << ++kase << ": maximum height = " << solve() << endl;
86     }
87 
88     return 0;
89 }