Matrix Continuous Product ZOJ 1276 Optimal Array multiplication Sequence

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1 /*2 Test Instructions: Add appropriate parentheses to change the order of calculations to minimize the total number of calculations3 matrix multiplication problem, DP solution: state transition equation: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] Record the location of the break (that is, the position of the parentheses), the backtracking output results6 */7#include <cstdio>8#include <cstring>9#include <string>Ten#include <algorithm> One#include <cmath> A#include <iostream> - using namespacestd; -  the Const intMAXN = 1e2 +Ten; - Const intINF =0x3f3f3f3f; - intDP[MAXN][MAXN]; - intS[MAXN][MAXN]; + intP[MAXN]; - intN; +  A voidPrintintIintj) at { -     if(i = = j) printf ("a%d", i); -     Else -     { -printf ("("); - print (i, s[i][j]); inprintf ("x"); -Print (S[i][j] +1, j); toprintf (")"); +     } - } the  * voidWorkvoid) $ {Panax Notoginseng      for(intI=1; i<=n; ++i) Dp[i][i] =0; -      for(intL=2; l<=n; ++l) the     { +          for(intI=1; i<=n-l+1; ++i) A         { the             intj = i + L-1; +DP[I][J] =INF; -              for(intK=i; k<=j-1; ++k) $             { $                 intTMP = Dp[i][k] + dp[k+1][J] + p[i-1] * P[k] *P[j]; -                 if(TMP <Dp[i][j]) -                 { theDP[I][J] = tmp; S[I][J] =K; -                 }Wuyi             } the         } -     } Wu  -Print (1, n); Puts (""); About } $  - intMainvoid)//ZOJ 1276 Optimal Array multiplication Sequence - { -     //freopen ("zoj_1276.in", "R", stdin); A  +     intCAS =0; the      while(SCANF ("%d", &n) = =1) -     { $         if(n = =0) Break; the          for(intI=1; i<=n; ++i) scanf ("%d%d", &p[i-1], &p[i]); the  theprintf ("Case %d:", ++CAs); the Work (); -     } in  the     return 0; the } About  the /* the Case 1: (A1 x (A2 x A3)) the Case 2: ((A1 x A2) x A3) + Case 3: ((A1 x (A2 x A3)) x ((A4 x A5) x A6)) - */

Matrix Continuous Product ZOJ 1276 Optimal Array multiplication Sequence