hdu 5578 Friendship of Frog

最新推荐文章于 2021-08-23 23:19:17 发布

原创最新推荐文章于 2021-08-23 23:19:17 发布 · 429 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一个算法问题，即计算一组代表不同国家青蛙的字符串中，相同字符（代表来自同一国家的青蛙）之间的最短距离。如果不存在相同的字符，则输出-1。文章提供了完整的代码实现，并通过样例输入和输出来验证算法的有效性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

N frogs from different countries are standing in a line. Each country is represented by a lowercase letter. The distance between adjacent frogs (e.g. the 1st and the 2nd frog, the N−1th and the Nth frog, etc) are exactly 1

. Two frogs are friends if they come from the same country.

The closest friends are a pair of friends with the minimum distance. Help us find that distance.

Input

First line contains an integer T, which indicates the number of test cases.

Every test case only contains a string with length N, and the ith character of the string indicates the country of ith frogs.

⋅ 1≤T≤50.

⋅ for 80% data, 1≤N≤100.

⋅ for 100% data, 1≤N≤1000.

⋅

the string only contains lowercase letters.

Output

For every test case, you should output " Case #x: y", where x indicates the case number and counts from 1 and y is the result. If there are no frogs in same country, output −1

instead.

Sample Input

2
abcecba
abc

Sample Output

Case #1: 2
Case #2: -1

这个就是求两个相同字母之间的最小距离，如果不纯在相同字母输出-1

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f

using namespace std;

char s[1100];

int main()
{
	ios::sync_with_stdio(false);
	int t, n, i, j, sum, x = 1;
	int m;
	cin>>t;
	while(t--)
	{
		cin>>s;
		n = strlen(s);
		m = inf;
		for(i = 0;i < n;i++)
		{
			sum = 0;
			for(j = i + 1;j < n;j++)
			{
				sum++;
				if(s[i] == s[j])
				{
					m = min(m,sum);
					break;
				}
			}
		}
		cout<<"Case #"<<x<<": ";
		x++;
		if(m < inf)
			cout<<m<<endl;
		else
			cout<<-1<<endl;
	}
	return 0;
}