ZOJ 3829 Known Notation (2014牡丹江H题)

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5383

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

Submit    Status

最近忙着各种事情，好久没写博客了。。。

这道题是牡丹H题，当时一直纠结一道概率DP。。。这题并没有深入的想，到比赛后才搞出来。。。

首先我们可以先对字符串进行补全，因为数字数目必须是’*‘数目+1。之后呢，我们发现对于‘*’，我们

其实只要把它尽可能的丢到后面去就完了。。然后从头到尾扫一遍。

不过有些特殊情况记得要小心。。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<bitset>
#include<cstdlib>
#define CLR(A) memset(A,0,sizeof(A))
using namespace std;
int main(){
    int T;
    while(~scanf("%d",&T)){
        while(T--){
            string str;
            cin>>str;
            int cnum=0,cop=0,ans=0;
            for(int i=0;str[i];i++){
                if(str[i]>='0'&&str[i]<='9') cnum++;
                else cop++;
            }
            if(cnum<cop+1) {
                ans+=(cop+1-cnum);
                cnum=cop+1-cnum;
            }
            else cnum=0;

            if(cop==0){
                cout<<"0"<<endl;
                continue;
            }
            int len=str.size();
            int last=len-1;
            bool flag=0;
            while(last>=0&&!(str[last]>='0'&&str[last]<='9')) last--;
            for(int i=0;str[i];i++){
                if(i>=last) break;
                if(str[i]>='0'&&str[i]<='9'){ cnum++;}
                else{
                    if(cnum<2){
                        ans++;
                        cnum++;
                        last--;
                        flag=1;
                        while(last>=0&&!(str[last]>='0'&&str[last]<='9')) last--;
                    }
                    else cnum--;
                }
            }
            if(flag==0&&str[len-1]!='*') ans++;
            cout<<ans<<endl;
        }
    }
    return 0;
}