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 */