【习题 6-9 UVA - 127】"Accordian" Patience

在这里输入题意

【题解】

链表模拟即可。
1pile不能加s...

【代码】

#include <bits/stdc++.h>

using namespace std;

const int N = 60;

string s[N];

int l[N], r[N];

vector <string> v[N];

int main() {

	#ifdef LOCAL_DEFINE

		freopen("F:\\c++source\\rush_in.txt", "r", stdin);

	#endif

	ios::sync_with_stdio(0), cin.tie(0);

	while (cin >> s[1] && s[1] != "#") {

		for (int i = 2; i <= 52; i++) cin >> s[i];

		for (int i = 1; i <= 52; i++) v[i].clear();

		for (int i = 1; i <= 52; i++) v[i].push_back(s[i]);

		for (int i = 0; i <= 52; i++) l[i] = i - 1, r[i] = i + 1;

		bool ok = true;

		while (ok) {

			ok = false;

			for (int i = r[0]; i != 53; i = r[i]) {

				int x1 = -1, x2 = -1, x3 = l[i];

				if (x3 >= 0) x2 = l[x3];

				if (x2 >= 0) x1 = l[x2];

				if (x1 >= 1 && x1 <= 52 && (v[x1].back()[0] == v[i].back()[0] || v[x1].back()[1] == v[i].back()[1])) {

					v[x1].push_back(v[i].back());

					v[i].pop_back();

					ok = true;

				}else

					if (x3 >= 1 && x3 <= 52 && (v[x3].back()[0] == v[i].back()[0] || v[x3].back()[1] == v[i].back()[1])) {

						v[x3].push_back(v[i].back());

						v[i].pop_back();

						ok = true;

					}

				if (v[i].empty()) {

					int ll = l[i], rr = r[i];

					r[ll] = rr;

					l[rr] = ll;

				}

				if (ok) break;

			}

		}

		vector <int> ans;

		ans.clear();

		for (int i = r[0]; i != 53; i = r[i]) {

			ans.push_back(v[i].size());

		}

		cout << (int)ans.size() << " pile";

		if ((int)ans.size()!=1) cout <<"s";

		cout<<" remaining:";

		for (int i = 0; i <(int)ans.size(); i++)

			cout << ' ' << ans[i];

		cout << endl;

	}

	return 0;

}

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【习题 6-9 UVA - 127】"Accordian" Patience

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