HDU-5578 Friendship of Frog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1134    Accepted Submission(s): 723

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 2ndfrog, 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

题意：

有不同国家的蛤，求相同国家相邻最近的两个蛤的距离，没有相等就输出-1.

上海的题，上海的题吧，果然是上海的题吧！！

共有26个字母 所以只要每一位向上比较26位就好了，最先出现的哦哦诶就是当前元素的最小值，注意不要越界。

附AC代码：

#include<bits/stdc++.h>
using namespace std;

const int INF=1<<30;


int main(){
    int t;
    cin>>t;
    int ans=1;
    while(t--){
        string s;
        cin>>s;
        int len=s.size();
        int MIN=INF;
        for(int i=0;i<len;i++){
            for(int j=1;j<=26&&i+j<len;j++){
                if(s[i]==s[i+j]){
                    MIN=min(MIN,j);
                    break;
                }
            }
        }
        if(MIN==INF)
        cout<<"Case #"<<ans++<<": -1"<<endl;
        else
        cout<<"Case #"<<ans++<<": "<<MIN<<endl;
    }
    return 0;
}

看到有大神用字符转换做，给跪：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[1010];
int a[30];
int main()
{
    int t,T=1,i,l;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,-1,sizeof(a));
        scanf("%s",s);
        l=strlen(s);
        int ans=0x3f3f3f;
        for(i=0;i<l;i++)
        {
            if(a[s[i]-'a']!=-1)
                ans=min(ans,i-a[s[i]-'a']);
            a[s[i]-'a']=i;
        }
        if(ans==0x3f3f3f)
            ans=-1;
        printf("Case #%d: %d
",T++,ans);
    }
    return 0;
}