【习题 5-9 UVA - 1596】Bug Hunt

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;

}

巴特西

【习题 5-9 UVA - 1596】Bug Hunt

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