ZOJ3829 ACM-ICPC 2014 亞洲地區賽牡丹江賽區現場賽K題 Known Notation 貪心，zoj3829acm-icpc

ZOJ3829 ACM-ICPC 2014 亞洲地區賽牡丹江賽區現場賽K題 Known Notation 貪心，zoj3829acm-icpc

Known NotationTime Limit: 2 Seconds      Memory Limit: 131072 KB

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

31*111*234***

Sample Output

102

    這道題是一個其實很簡單，然後WA點很多的題目，注意便是極好的~~~，題目意思是給你一個式子，問你變成尾碼運算式需要的最少步驟，題目解法就是很簡單的，貪心，對於一個尾碼運算式，必須滿足符號數量小於數字數量，那麼缺幾個數字補幾個，那麼就是insert的主要步驟，然後，對於運算式，最後一位如果是數字，並且沒有補充星號，則需要與前面星號交換位置，注意這些以後，就直接掃一遍，棧內初始化數量為之前步驟添加的數字數，沒有添加就為0，數字就入棧入棧一個，符號就出棧一個，當數字數量少於2時了就交換後面的數字，然後計算交換和插入總數輸出即可AC，具體AC代碼如下：

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>#include<map>#include<vector>#include<queue>using namespace std;char ch[1005];int main(){  //  freopen("in.txt","r",stdin);    int t;    cin>>t;    while(t--)    {        cin>>ch;        int len=strlen(ch);        int sum1=0,sum2=0;        for(int i=0;i<len;i++)        {            if(ch[i]=='*')                sum1++;            else                sum2++;        }        bool f=false;        if(ch[len-1]=='*')            f=true;        int res=0;        sum1++;        int pas=0;        if(sum1>sum2)            res=sum1-sum2;        else            pas=sum2-sum1,f=true;        int sum=res;        if(f==false)        {            int i;            for(i=0;i<len;i++)                if(ch[i]=='*')                    break;            swap(ch[len-1],ch[i]);            res++;        }        for(int i=0;i<len;i++)        {            if(ch[i]!='*')                sum++;            else            {                if(sum<2)                {                    res++;                    if(pas>0)                        pas--;                    else                    {                        int j;                        for(j=len-1;j>=0;j--)                        {                            if(ch[j]!='*')                                break;                        }                        swap(ch[i],ch[j]);                    }                    sum++;                }                else                    sum--;            }        }        printf("%d\n",res);    }    return 0;}