2014ACM/ICPC亞洲地區賽牡丹江站現場賽-K （ ZOJ 3829 ） Known Notation，2014acmicpc

2014ACM/ICPC亞洲地區賽牡丹江站現場賽-K （ ZOJ 3829 ） Known Notation，2014acmicpc

Known NotationTime Limit: 2 Seconds      Memory Limit: 65536 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

Author: CHEN, Cong
Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

題目連結：Known Notation

解題思路：貪心。如果num < star 時，則必須在前面補充  star - num + 1  個數字，因為star個星星，需要star+1個數字，才符合要求。接下來，盡量把數字放到前面，把星星放到後面，兩個數字可以消掉一個星星，因為這時候 a*b 相當於一個數字了。如果前面的數字不夠用，就用前面的星星和後面的數字交換，因為交換比插入的結果要好。不斷貪心下去，即可。

AC代碼：

#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 0x7fffffffint 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]);     //前面的數字不夠，用前面的星星和後面的數字交換                    ans ++;                    p --;                    left_num += 2;                }            }            else left_num ++;        }        cout<<ans<<endl;    }    return 0;}