uva442 Matrix Chain Multiplication

题意：给定n个矩阵 然后给出每个矩阵的行数和列数 让你求出一个矩阵表达式的乘的次数

解题思路：栈

解题代码：

// File Name: uva442.c
// Author: darkdream
// Created Time: 2013年05月15日 星期三 21时15分38秒

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
struct node 
{
  int a, b;
}m[300] ;
int main(){

   //freopen("/home/plac/problem/input.txt","r",stdin);
   //freopen("/home/plac/problem/output.txt","w",stdout);
   int n; 
   scanf("%d",&n);
   char str[1000];
   getchar();
   for(int i = 1;i <= n;i ++)
   {
    char c;
    c = getchar();
    scanf("%d %d",&m[c-'A'+1].a,&m[c-'A'+1].b);
    getchar();
   }
   int a[1000];
   int b[1000] ;
   while(scanf("%s",str) != EOF)
   {
     int sum = 0 ,k = n+1, ok = 1;
     struct node temp;
     if(strlen(str) == 1)
         printf("0\n");
     else
     {
         for(int i = 0 ;i < strlen(str); i ++)
         {
            if(str[i] == '(')
                a[i] = -1 ;
            else if(str[i] == ')')
                a[i] = -2;
            else
                a[i] = str[i]-'A' +1;
         }
         int j = -1 ;
         for(int  i = 0 ;i <strlen(str) ; i++)
         {
            if(a[i] == -1)
            {
               j++ ;
               b[j] = a[i];
            }
            else if(a[i] == -2)
            {
                b[j-1] = b[j];
                b[j] = 0 ;
                j--;
                if(b[j-1] > 0)
                {
                 if(m[b[j]].a == m[b[j-1]].b)
                    {
                        sum += m[b[j-1]].a*m[b[j-1]].b*m[b[j]].b;
                        m[k].a = m[b[j-1]].a;
                        m[k].b = m[b[j]].b;
                        b[j-1] = k;
                        b[j] = 0;
                        j--;
                        k++;
                    
                    }
                 else
                  {
                    ok = 0 ;
                    break;
                  }
                   
                }

            }
            else
            {
              if(b[j] > 0 )
              {
                 if(m[a[i]].a == m[b[j]].b)
                    {
                        sum += m[b[j]].a*m[b[j]].b*m[a[i]].b;
                        m[k].a = m[b[j]].a;
                        m[k].b = m[a[i]].b;
                        b[j] = k;
                        k++;
                    
                    }
                 else
                  {
                    ok = 0 ;
                    break;
                  }
              }
              else
              {
                j++;
                b[j] = a[i];
              }
            }
         }

     if(ok)
         printf("%d\n",sum);
     else
         printf("error\n");
     }

   }

   
return 0 ;
}