HDU 5578 Friendship of Frog(暴力)——2015ACM/ICPC亚洲区上海站

Friendship of Frog

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0

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
   
   
   
   

 

/*********************************************************************/

题意：给你一个字符串,找最近的相同字符

解题思路：最多才1000个字符,既然是签到题,就不要多想了,暴力上就可以了

对于第i个字符,往后找相同字符,不断更新Min值就ok了

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
#define MAX(a,b) ((a)>(b)?(a):(b))
using namespace std;
const int N = 1005;
const int M = 2010;
const int inf = 2147483647;
const int mod = 2009;
char s[N];
int main()
{ 
    int t,i,j,Min,p=1;
    scanf("%d",&t);
    while(t--)
    {
        Min=inf;
        scanf("%s",s);
        for(i=0;s[i]!='\0';i++)
            for(j=i+1;s[j]!='\0';j++)
                if(s[i]==s[j])
                    Min=min(Min,j-i);
        printf("Case #%d: ",p++);                    
        if(Min!=inf)
            printf("%d\n",Min);
        else
            puts("-1");
    }
    return 0; 
}

菜鸟成长记