矩阵连乘积 ZOJ 1276 Optimal Array Multiplication Sequence

 

题目传送门

 1 /*
 2     题意：加上适当的括号，改变计算顺序使得总的计算次数最少
 3     矩阵连乘积问题，DP解决：状态转移方程：
 4     dp[i][j] = min (dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j])    (i<=k<j)
 5     s[i][j] 记录断开的地方(即加括号的位置)，回溯法输出结果
 6 */
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <string>
10 #include <algorithm>
11 #include <cmath>
12 #include <iostream>
13 using namespace std;
14 
15 const int MAXN = 1e2 + 10;
16 const int INF = 0x3f3f3f3f;
17 int dp[MAXN][MAXN];
18 int s[MAXN][MAXN];
19 int p[MAXN];
20 int n;
21 
22 void print(int i, int j)
23 {
24     if (i == j)    printf ("A%d", i);
25     else
26     {
27         printf ("(");
28         print (i, s[i][j]);
29         printf (" x ");
30         print (s[i][j] + 1, j);
31         printf (")");
32     }
33 }
34 
35 void work(void)
36 {
37     for (int i=1; i<=n; ++i)    dp[i][i] = 0;
38     for (int l=2; l<=n; ++l)
39     {
40         for (int i=1; i<=n-l+1; ++i)
41         {
42             int j = i + l - 1;
43             dp[i][j] = INF;
44             for (int k=i; k<=j-1; ++k)
45             {
46                 int tmp = dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j];
47                 if (tmp < dp[i][j])
48                 {
49                     dp[i][j] = tmp;    s[i][j] = k;
50                 }
51             }
52         }
53     }
54 
55     print (1, n);    puts ("");
56 }
57 
58 int main(void)        //ZOJ 1276 Optimal Array Multiplication Sequence
59 {
60     //freopen ("ZOJ_1276.in", "r", stdin);
61 
62     int cas = 0;
63     while (scanf ("%d", &n) == 1)
64     {
65         if (n == 0)    break;
66         for (int i=1; i<=n; ++i)    scanf ("%d%d", &p[i-1], &p[i]);
67 
68         printf ("Case %d: ", ++cas);
69         work ();
70     }
71 
72     return 0;
73 }
74 
75 /*
76 Case 1: (A1 x (A2 x A3))
77 Case 2: ((A1 x A2) x A3)
78 Case 3: ((A1 x (A2 x A3)) x ((A4 x A5) x A6))
79 */

 

转载于:https://www.cnblogs.com/Running-Time/p/4490772.html