[Greedy + simulation] zoj 3829 known notation

Question link:

Http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 5383

Known notation Time 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 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

Author: Chen, Cong
Submit status Question meaning:

1 ~ A string consisting of 9 and *. You can add a number or * in two ways to exchange any two characters. The formula used to calculate the minimum number of steps is inverse polish.

Solution:
Greedy + Simulation

1. When the number of numbers is less than or equal to *, add a number to set the number.

2. In the case of 111*111, the answer is 1. Record the minimum number and the maximum number of current numbers. * The two parts are separated, and numbers cannot be connected together. Mi must be added separately, ma.

3. If the preceding number is not enough, the last number is exchanged with the current *. Note that after the number is exchanged, it can be connected to the preceding number. Therefore, mi cannot be added.

Code:

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 1100char sa[Maxn];int n;int main(){    //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   int t;   scanf("%d",&t);   while(t--)   {       scanf("%s",sa+1);       n=strlen(sa+1);       int cnt1=0,cnt2=0;       int la=0,Ma,Mi,p,ans=0;       for(int i=1;i<=n;i++)       {           if(sa[i]=='*')                cnt2++;           else                cnt1++;       }       if(cnt1<=cnt2)       {           ans+=cnt2+1-cnt1;           la=cnt2+1-cnt1;       }       for(p=1;p<=n&&sa[p]!='*';p++)       if(p>n)       {           printf("0\n");           continue;       }       la+=p-1;       Ma=la;       if(la)  //注意这里区别一开始都是*的情况            Mi=1;       else            Mi=0;       //printf("Mi:%d Ma:%d la:%d p:%d\n",Mi,Ma,la,p);       //system("pause");       for(;p<=n;p++)       {           if(sa[p]=='*')           {               if(Ma<=1) //非得交换               {                   ans++;                   for(int j=n;j>p;j--)                   {                       if(sa[j]!='*')//最后一定有数字，一定可以交换                       {                           swap(sa[p],sa[j]);                           break;                       }                   }                   int temp=0;                   while(p<=n&&sa[p]!='*')                        temp++,p++;                   Ma+=temp;                   if(!Mi)                        Mi=1;  //为第一个  否则不能把Mi加一个，交换后可以连在一起的                   p--;               }               else  //               {                   if(Mi>1)                        Mi--;                   Ma--;               }           }           else           {               int temp=0;               while(p<=n&&sa[p]!='*')                   temp++,p++;               Mi++;  //注意*隔开的两部分，不能连在一起，所以Mi要加一               Ma+=temp;               p--;           }       }       ans+=Mi-1;       printf("%d\n",ans);   }    return 0;}/*111*1111*11*11*/

[Greedy + simulation] zoj 3829 known notation