【习题 5-9 UVA - 1596】Bug Hunt

【链接】 我是链接,点我呀:)

【题意】

在这里输入题意

【题解】

map模拟
map记录每个数组的大小
map ,int>记录数组的某个下标的值。
递归处理嵌套的情况就好

【代码】

#include <bits/stdc++.h>
using namespace std;

string s;
map <string, int> mmap;
map <pair<string, int> ,int> mmap2;
int cur = 0;
bool ok;
int gl(int l, int r)
{
	return r - l + 1;
}

pair <string, string> cl(string s)
{
	string val = "", temp = "";
	int now = 0;
	while (s[now] != '[') val += s[now++];
	int len = s.size();
	temp = s.substr(now + 1, gl(now + 1, len - 2));
	return make_pair(val, temp);
}

int get_num(string s)
{
	if (isdigit(s[0]))
	{
		int x = 0, lens = s.size();
		for (int i = 0; i < lens; i++)
			x = x * 10 + s[i] - '0';
		return x;
	}
	string val = "";
	int len = s.size(),l;
	for (int i = 0; i < len; i++)
	{
		if (s[i] == '[')
		{
			l = i;
			break;
		}
		val += s[i];
	}
	int nex = get_num(s.substr(l + 1, gl(l + 1, len - 2)) );
	if (nex >= mmap[val] || nex < 0)
	{
		ok = false;
		return -1;
	}
	else
	{
		if (mmap2.find(make_pair(val, nex)) != mmap2.end())
		{
			return mmap2[make_pair(val, nex)];
		}
		else
		{
			ok = false;
			return -1;
		}
	}
}

int main()
{
	//freopen("F:\\rush.txt", "r", stdin);
	while (cin >> s && s[0] != '.')
	{
		mmap.clear();
		mmap2.clear();
		ok = true;
		cur = 1;
		while (s[0] != '.')
		{
			if (!ok)
			{
				cin >> s;
				continue;
			}
			int fi = s.find('=', 0);
			if (fi != -1)// a=b
			{
				string s1 = s.substr(0, fi);
				string s2 = s.substr(fi + 1);

				pair <string, string> temp = cl(s1);
				s1 = temp.second;
				string a = temp.first;

				int x = get_num(s1), y = get_num(s2);
				if (!ok || x < 0 || x >= mmap[a])
				{
					ok = 0;
					cout << cur << endl;
					continue;
				}
				mmap2[make_pair(a, x)] = y;
			}
			else
			{
				pair <string, string> temp = cl(s);
				mmap[temp.first] = get_num(temp.second);
			}

			cur++;
			cin >> s;
		}
		if (ok) cout << 0 << endl;
	}
	return 0;
}