Codeforces Round #617 (Div. 3) C. Yet Another Walking Robot

http://codeforces.com/contest/1296/problem/C

题意:给一段字符串表示移动,然后求删除最短的一段,并且不影响结果

题解:

意思是:建立pair点和map,当遍历到第i个点有一个pair值,把这个加到map里面,如果向后接着遍历时出现与i点相同的pair值时,那么这一段表示可以删除的一段

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif

	int t;
	cin >> t;
	while (t--) {
		int n;
		string s;
		cin >> n >> s;
		int l = -1, r = n;
		map<pair<int, int>, int> vis;
		pair<int, int> cur = {0, 0};
		vis[cur] = 0;
		for (int i = 0; i < n; ++i) {
			if (s[i] == ‘L‘) --cur.first;
			if (s[i] == ‘R‘) ++cur.first;
			if (s[i] == ‘U‘) ++cur.second;
			if (s[i] == ‘D‘) --cur.second;
			if (vis.count(cur)) {
				if (i - vis[cur] + 1 < r - l + 1) {
					l = vis[cur];
					r = i;
				}
			}
			vis[cur] = i + 1;
		}
		if (l == -1) {
			cout << -1 << endl;
		} else {
			cout << l + 1 << " " << r + 1 << endl;
		}
	}

	return 0;
}

原文地址：https://www.cnblogs.com/RE-TLE/p/12267437.html