ACM learning process -- zoj 3829 known notation (2014 Mudanjiang semi-finals K questions) (strategy, stack)

Description

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 learned it before, we will offer some examples here. for instance, to add 3 and 4, one wocould 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 wocould 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 integerTIndicating 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

31*111*234***

Sample output

102


It is not particularly difficult to think of matching strategies for this question.
To solve this problem, we must first determine that the number of numbers must be greater than the number of operators. Because this is a binary operation, two numbers and an operator can generate a number. Secondly, this operation method has a combination law. Therefore, when multiple numbers correspond to one operator, you can select a certain number of numbers and operators nearby for calculation.
In this case, when the number of operators is greater than or equal to the number of numbers, the insert number operation can be prioritized so that all the inserted numbers are inserted at the beginning, because of the combination law, this is the best case. Second, you only need to satisfy the following conditions before each operator: up to now, the number of operators is always smaller than the number of numbers. Otherwise, the number at the end of the current operator is exchanged with the current operator (due to the relationship of the combination law, consider the last number ).
Finally, we need to consider two issues:
1. In the beginning, it was a pure number question. The answer should be 0;
2. After all the operations are completed, the last character is not *. The answer should be added.


Code:

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <set>#include <map>#include <vector>#include <string>#include <queue>#define inf 0x3fffffff#define esp 1e-10using namespace std;int numc, numi, len;char s[1005];int Stack[1005], top;void Input (){    numc = 0;    numi = 0;    len = 0;    top = 0;    char ch;    for (;;)    {        ch = getchar();        if (ch == ‘\n‘)            break;        if (ch == ‘*‘)            numc++;        else        {            numi++;            Stack[top++] = len;        }        s[len++] = ch;    }}int qt(){    if (numc == 0)        return 0;    int ans = 0;    if (numi > numc)        numi = numc = 0;    else    {        ans = numi = numc - numi + 1;        numc = 0;    }    for (int i = 0; i < len; ++i)    {        if (s[i] == ‘*‘)            numc++;        else            numi++;        if (numc >= numi)        {            swap (s[i], s[Stack[top-1]]);            top--;            ans++;            numc--;            numi++;        }    }    if (s[len-1] != ‘*‘)        ans++;    return ans;}int main(){    //freopen ("test.txt", "r", stdin);    int T;    scanf ("%d", &T);    getchar();    for (int times = 0; times < T; ++times)    {        Input();        printf ("%d\n", qt());    }    return 0;}

 

ACM learning process -- zoj 3829 known notation (2014 Mudanjiang semi-finals K questions) (strategy, stack)