Bug Hunt UVA - 1596

  In this problem, we consider a simple programming language that has only declarations of onedimensional integer arrays and assignment statements. The problem is to find a bug in the given program. The syntax of this language is given in BNF as follows:

  where ⟨new line⟩ denotes a new line character (LF).

  Characters used in a program are alphabetical letters, decimal digits, =, [, ] and new line characters. No other characters appear in a program.

  A declaration declares an array and specifies its length. Valid indices of an array of length n are integers between 0 and n − 1, inclusive. Note that the array names are case sensitive, i.e. array a and array A are different arrays. The initial value of each element in the declared array is undefined.

  For example, array a of length 10 and array b of length 5 are declared respectively as follows.

a[10] b[5]

  An expression evaluates to a non-negative integer. A ⟨number⟩ is interpreted as a decimal integer. An ⟨array name⟩ [⟨expression⟩] evaluates to the value of the ⟨expression⟩-th element of the array. An assignment assigns the value denoted by the right hand side to the array element specified by the left hand side.

Examples of assignments are as follows.

a[0]=3 a[1]=0 a[2]=a[a[1]] a[a[0]]=a[1]

  A program is executed from the first line, line by line. You can assume that an array is declared once and only once before any of its element is assigned or referred to.

  Given a program, you are requested to find the following bugs.

• An index of an array is invalid.

• An array element that has not been assigned before is referred to in an assignment as an index of array or as the value to be assigned.

  You can assume that other bugs, such as syntax errors, do not appear. You can also assume that integers represented by ⟨number⟩s are between 0 and 231 − 1 (= 2147483647), inclusive.

Input

  The input consists of multiple datasets followed by a line which contains only a single ‘.’ (period).

  Each dataset consists of a program also followed by a line which contains only a single ‘.’ (period).

  A program does not exceed 1000 lines. Any line does not exceed 80 characters excluding a new line character.

Output

  For each program in the input, you should answer the line number of the assignment in which the first bug appears. The line numbers start with 1 for each program. If the program does not have a bug, you should answer zero. The output should not contain extra characters such as spaces.

Sample Input

a[3]
a[0]=a[1]
.
x[1]
x[0]=x[0]
.
a[0]
a[0]=1
.
b[2]
b[0]=2
b[1]=b[b[0]]
b[0]=b[1]
.
g[2]
G[10]
g[0]=0
g[1]=G[0]
.
a[2147483647]
a[0]=1
B[2]
B[a[0]]=2
a[B[a[0]]]=3
a[2147483646]=a[2]
.
a[32]
b[2]
b[1]=33
b[1]=30
a[b[1]]=15
.
.

Sample Output

2
2
2
3
4
0
0

HINT

此题目的细节较多，首先补充一个测试样例：

a[32]
b[2]
b[1]=33
b[1]=30
a[b[1]]=15

0		//输出结果

这个样例的目的是检测你的程序有没有对已经赋值的元素重新正确赋值的能力，样例里面并没有，这个bug浪费了1个小时。

下面一步一步的分析：

1. 输入的样例程序只可能右两中错误，即：数组下标越界，使用未经赋值的元素（可能同时出现在一个赋值语句中）。

2. 所有可能出现的样例程序语句有:

   a[32]	//声明数组语句
   a[2]=1	//赋值语句
   a[a[2]]=a[2]//下标嵌套元素值
   a[a[2]]=3//	对元素的值重新赋值
   

3. 如何区分声明语句以及赋值或者修改值语句？

   使用find()查找有没有‘=’，有等于号就不是声明语句。

4. 如何存储数据？

   对数组的名以及边界采用一个map来存储，对赋值或者初始化的元素的值采用另一个map来存储，这个map的键为数组名称，值为一个map，里面存储此数组的每一个元素的值。

5. 如何处理数据？

   对于一个正确的声明语句，第一个字母是数组的名称，最外层[]以内的部分最终一定可以计算化为一个数字。

   对于一个正确的赋值语句，等号右边一定可以化为一个数字；等号右边类似于声明语句，最外层[]以内的部分最终一定可以计算化为一个数字。

   因此，上面的讨论得到的区别就是，有没有最外层[]的问题。因此我们对[]进行单独讨论，剩下的就是可以化简称为一个数组的部分。

   对于化简成为数字的部分的操作就是栈操作，剩下的就是一些计算下标值和判断了。

Accepted

#include<bits/stdc++.h>
using namespace std;
map<int,unsigned long int>arrname;										//保存数组名以及下标边界
map<int, map<unsigned long int, unsigned long int> >arrvalue;			//保存已经赋值的元素
int transfor(string s, int flag,int &value) {							//计算
	value = 0;															//初始化
	stack<char>name;													//保存数组名称
	for (int i = 0;i < s.size();i++) {									//对每一个字符判断处理
		if (isalpha(s[i]))name.push(s[i]);								//如果是数组名，入栈
		if (s[i] >= '0' && s[i] <= '9')value = value * 10 + s[i] - '0';	//如果是数字，计算
		if (s[i] == ']') {												//如果是]出栈
			if (!arrname.count((int)name.top())||arrname[(int)name.top()] < value || !arrvalue[(int)name.top()].count(value)) {								//如果数组未声明或者下标越界或者未进行初始化的元素
				if (!arrname.count((int)name.top()) && name.size() == 1 && flag == 2) {
					arrname[(int)name.top()] = value;					//如果他是声明语句，则声明入栈
					return 1;
				}
				return 0;												//否则，返回错误
			}
			value = arrvalue[(int)name.top()][value];					//如果一切正常，则计算新的下标
			name.pop();
		}
	}
	return 1;
}
int main() {

	string s1, s2, s;
	int num = 0, flag = 1,value1,value2;
	while (cin >> s) {
		if (s == ".")break;
		num = 0;arrname.clear();arrvalue.clear();flag = 1;				//初始化
		while (s != ".") {
			num++;
			int i = s.find('=');										//判断是不是一个声明语句
			if (i != - 1) {												//是赋值语句
				s1 = s.substr(2, i-3);									//将左值最外侧最外层的[]去除
				s2 = s.substr(i + 1);
				if (!transfor(s1, 1, value1) || !transfor(s2, 0, value2)|| value1 >= arrname[(int)s[0]]) {flag = 0;break;}														//判断是否是否合法
				else arrvalue[s[0]][value1] = value2;					//合法就赋值
			}
			else if (!transfor(s, 2,value1)) {flag = 0;break;}			//声明是否合法
			cin >> s;
		}
		if (!flag) {													//结果输出
			cout << num << endl;
			while (s != ".")cin >> s;
		}
		else cout << 0 << endl;
	}
}