矩阵连乘积 ZOJ 1276 Optimal Array Multiplication Sequence

题目传送门

 /*
     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
     矩阵连乘积问题，DP解决：状态转移方程：
     dp[i][j] = min (dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j])    (i<=k<j)
     s[i][j] 记录断开的地方(即加括号的位置)，回溯法输出结果
 */
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <cmath>
 #include <iostream>
 using namespace std;

 const int MAXN = 1e2 + ;
 const int INF = 0x3f3f3f3f;
 int dp[MAXN][MAXN];
 int s[MAXN][MAXN];
 int p[MAXN];
 int n;

 void print(int i, int j)
 {
     if (i == j)    printf ("A%d", i);
     else
     {
         printf ("(");
         print (i, s[i][j]);
         printf (" x ");
         print (s[i][j] + , j);
         printf (")");
     }
 }

 void work(void)
 {
     for (int i=; i<=n; ++i)    dp[i][i] = ;
     for (int l=; l<=n; ++l)
     {
         for (int i=; i<=n-l+; ++i)
         {
             int j = i + l - ;
             dp[i][j] = INF;
             for (int k=i; k<=j-; ++k)
             {
                 int tmp = dp[i][k] + dp[k+][j] + p[i-] * p[k] * p[j];
                 if (tmp < dp[i][j])
                 {
                     dp[i][j] = tmp;    s[i][j] = k;
                 }
             }
         }
     }

     print (, n);    puts ("");
 }

 int main(void)        //ZOJ 1276 Optimal Array Multiplication Sequence
 {
     //freopen ("ZOJ_1276.in", "r", stdin);

     int cas = ;
     while (scanf ("%d", &n) == )
     {
         if (n == )    break;
         for (int i=; i<=n; ++i)    scanf ("%d%d", &p[i-], &p[i]);

         printf ("Case %d: ", ++cas);
         work ();
     }

     return ;
 }

 /*
 Case 1: (A1 x (A2 x A3))
 Case 2: ((A1 x A2) x A3)
 Case 3: ((A1 x (A2 x A3)) x ((A4 x A5) x A6))
 */