2014牡丹江K Known Notation

Known Notation

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Do you know reverse Polish notation (RPN)? It is a known notation in the area of mathematics and computer science. It is also known as postfix notation since every operator in an expression follows all of its operands. Bob is a student in Marjar University. He is learning RPN recent days.

To clarify the syntax of RPN for those who haven't learnt it before, we will offer some examples here. For instance, to add 3 and 4, one would write "3 4 +" rather than "3 + 4". If there are multiple operations, the operator is given immediately after its second operand. The arithmetic expression written "3 - 4 + 5" in conventional notation would be written "3 4 - 5 +" in RPN: 4 is first subtracted from 3, and then 5 added to it. Another infix expression "5 + ((1 + 2) × 4) - 3" can be written down like this in RPN: "5 1 2 + 4 × + 3 -". An advantage of RPN is that it obviates the need for parentheses that are required by infix.

In this problem, we will use the asterisk "*" as the only operator and digits from "1" to "9" (without "0") as components of operands.

You are given an expression in reverse Polish notation. Unfortunately, all space characters are missing. That means the expression are concatenated into several long numeric sequence which are separated by asterisks. So you cannot distinguish the numbers from the given string.

You task is to check whether the given string can represent a valid RPN expression. If the given string cannot represent any valid RPN, please find out the minimal number of operations to make it valid. There are two types of operation to adjust the given string:

 

1. Insert. You can insert a non-zero digit or an asterisk anywhere. For example, if you insert a "1" at the beginning of "2*3*4", the string becomes "12*3*4".
2. Swap. You can swap any two characters in the string. For example, if you swap the last two characters of "12*3*4", the string becomes "12*34*".

 

The strings "2*3*4" and "12*3*4" cannot represent any valid RPN, but the string "12*34*" can represent a valid RPN which is "1 2 * 34 *".

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is a non-empty string consists of asterisks and non-zero digits. The length of the string will not exceed 1000.

Output

For each test case, output the minimal number of operations to make the given string able to represent a valid RPN.

Sample Input

3
1*1
11*234**
*

Sample Output

1
0
2

看来只要认真思考，这些较为简单的题都是可以A的。

想了一个上午，一开始没把逆波兰表达式的可能性想清楚，导致算法错误，后来想到只要前面数字的个数大于*的个数，都是成立的，而且交换要比增加效率高。只有当数字个数小于*个数的时候才会增加数字，而且增加在最前面，这样是最优方案。之后只要发现到某一位*的个数大于等于数字的个数了，就把*和最后一个数字交换就可以了。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define M(a,b) memset(a,b,sizeof(a))
 7 typedef long long LL;
 8 using namespace std;
 9 
10 char num[1005];
11 
12 int main()
13 {
14     int t;
15     int ans = 0;
16     scanf("%d",&t);
17     while(t--)
18     {
19         scanf("%s",num);
20         int sl = strlen(num);
21         int cnt1 = 0;
22         int cnt2 = 0;
23         int flag = -1;
24         int bu = 0;
25         int seg = 0;
26         int step = 0;
27         int save = -1;
28         for(int i = sl-1;i>=0;i--)
29         {
30             if(num[i]=='*')
31               save = i;
32         }
33         if(num[sl-1] != '*')
34         {
35             if(save!=-1)
36               {swap(num[sl-1],num[save]);
37                step++;}
38         }
39         int tm1 = 0;
40         int tm2 = 0;
41         int ed;
42         for(int i = 0;i<sl;i++)
43         {
44             if(num[i]=='*') tm1++;
45             else tm2++;
46         }
47         if(tm1>=tm2) {
48                 step+=(tm1-tm2+1);
49          cnt2=(tm1-tm2+1);
50          ed = cnt2;
51         }
52         //cout<<ed<<endl;
53         for(int i = 0;i<sl;i++)
54         {
55             if(num[i]=='*')
56             {
57                 cnt1++;
58                 if(cnt1>=cnt2)
59                 {
60                     int te1,te2;
61                     for(int p = 0;p<sl;p++)
62                         if(num[p] != '*') te1 = p;
63                     for(int q = sl-1;q>=0;q--)
64                         if(num[q] == '*') te2 = q;
65                     swap(num[te1],num[te2]);
66                     step++;
67                     cnt1 = 0;
68                     cnt2 = ed;
69                     i = -1;
70                     continue;
71                 }
72             }
73             else cnt2++;
74         }
75         printf("%d\n",step);
76     }
77     return 0;
78 }

 

转载于:https://www.cnblogs.com/haohaooo/p/4025100.html