ACM學習曆程——ZOJ 3829 Known Notation （2014牡丹江地區賽K題）（策略，棧）

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Description

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


這個題目要能想到匹配的策略，其實還是不是特別難的。
對於這個題目，首先要確定的前提就是數位個數一定要大於運算子的個數，因為這是二元運算，故兩個數字和一個運算子能產生一個數字。其次，這種運演算法，是具備結合律的，所以，對於多個數字對應一個運算子的情況，可以就近選擇一定數量的數字和運算子進行運算。
這種情況下，當運算子數目大於等於數字數目時，可以優先考慮插入數的操作，讓插入的數全部插在最前面，因為結合律的緣故，這種情況是最優的。其次，只需要每個運算子前都滿足：到當前為止，運算子的個數始終小於數字個數，否則，將排在當前最後一個的數字與當前運算子交換（由於結合律的關係，考慮最後一個數字）。
最後，還要考慮兩個問題：
1、一開始就是純數位問題，這個答案應該為0；
2、執行完所有的操作，最後一個字元不是*的，答案應該再加1.


代碼：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <set>#include <map>#include <vector>#include <string>#include <queue>#define inf 0x3fffffff#define esp 1e-10using namespace std;int numc, numi, len;char s[1005];int Stack[1005], top;void Input (){    numc = 0;    numi = 0;    len = 0;    top = 0;    char ch;    for (;;)    {        ch = getchar();        if (ch == ‘\n‘)            break;        if (ch == ‘*‘)            numc++;        else        {            numi++;            Stack[top++] = len;        }        s[len++] = ch;    }}int qt(){    if (numc == 0)        return 0;    int ans = 0;    if (numi > numc)        numi = numc = 0;    else    {        ans = numi = numc - numi + 1;        numc = 0;    }    for (int i = 0; i < len; ++i)    {        if (s[i] == ‘*‘)            numc++;        else            numi++;        if (numc >= numi)        {            swap (s[i], s[Stack[top-1]]);            top--;            ans++;            numc--;            numi++;        }    }    if (s[len-1] != ‘*‘)        ans++;    return ans;}int main(){    //freopen ("test.txt", "r", stdin);    int T;    scanf ("%d", &T);    getchar();    for (int times = 0; times < T; ++times)    {        Input();        printf ("%d\n", qt());    }    return 0;}

 

ACM學習曆程——ZOJ 3829 Known Notation （2014牡丹江地區賽K題）（策略，棧）