2014ACM/ICPC亚洲区域赛牡丹江站现场赛-K （ ZOJ 3829 ） Known Notation

Known Notation

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

Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

题目链接：Known Notation

解题思路：贪心。如果num < star 时，则必须在前面补充  star - num + 1  个数字，因为star个星星，需要star+1个数字，才符合要求。接下来，尽量把数字放到前面，把星星放到后面，两个数字可以消掉一个星星，因为这时候 a*b 相当于一个数字了。如果前面的数字不够用，就用前面的星星和后面的数字交换，因为交换比插入的结果要好。不断贪心下去，即可。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    int n;
    string s;
    scanf("%d",&n);
    while(n--)
    {
        int num = 0, star = 0;
        cin>>s;
        int len = s.size();
        for(int i=0; i<len; i++){
            if(s[i] == '*') star ++;
            else num ++;
        }

        int left_num = 0, ans = 0;
        if(num < star){
            left_num += star - num + 1;
            ans += left_num;
        }

        for(int i=0, p = len-1; i<len; i++){
            while(i < p && s[p] == '*') p --;
            if(s[i] == '*'){
                left_num --;
                if(left_num < 1){
                    swap(s[i], s[p]);     //前面的数字不够，用前面的星星和后面的数字交换
                    ans ++;
                    p --;
                    left_num += 2;
                }
            }
            else left_num ++;
        }
        cout<<ans<<endl;
    }
    return 0;
}