HDU5578 Friendship of Frog

Friendship of Frog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1463    Accepted Submission(s): 960

Problem Description

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

 

Source

2015ACM/ICPC亞洲區上海站-重現賽（感謝華東理工）

題意：給一個長度不超過1000的字符串，求相同字符最小間隔，沒有相同字符輸出-1

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define ll __int64
using namespace std;

char s[2000];

int main()
{
    int T;
    scanf("%d",&T);
    int ca=1;
    while(T--)
    {

        scanf("%s",s);

        int len=strlen(s);
        int ans=999999;
        int flag;
        for(int i=0;i<26;i++)
        {
            int f='a'+i;
            flag=0;
           int pos;
           int dis=999999;
            for(int j=0;j<len;j++)
            {
                if(f==s[j])
                {
                    if(flag==0)//記錄該字符第一次出現的位置
                    {
                        pos=j;
                        flag=1;
                        continue;
                    }
                    dis=min(dis,j-pos);
                    pos=j;
                }
            }
                ans=min(dis,ans);
        }

        if(ans==999999)
            ans=-1;

        printf("Case #%d: %d\n",ca++,ans);

    }

    return 0;
}