HDU-5578 Friendship of Frog
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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1134 Accepted Submission(s): 723Problem DescriptionN 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.
InputFirst 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.
OutputFor 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 Input2abcecbaabcSample OutputCase #1: 2Case #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:
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int INF=1<<30; 5 6 7 int main(){ 8 int t; 9 cin>>t; 10 int ans=1; 11 while(t--){ 12 string s; 13 cin>>s; 14 int len=s.size(); 15 int MIN=INF; 16 for(int i=0;i<len;i++){ 17 for(int j=1;j<=26&&i+j<len;j++){ 18 if(s[i]==s[i+j]){ 19 MIN=min(MIN,j); 20 break; 21 } 22 } 23 } 24 if(MIN==INF) 25 cout<<"Case #"<<ans++<<": -1"<<endl; 26 else 27 cout<<"Case #"<<ans++<<": "<<MIN<<endl; 28 } 29 return 0; 30 }
看到有大神用字符转换做,给跪:
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 char s[1010]; 6 int a[30]; 7 int main() 8 { 9 int t,T=1,i,l; 10 scanf("%d",&t); 11 while(t--) 12 { 13 memset(a,-1,sizeof(a)); 14 scanf("%s",s); 15 l=strlen(s); 16 int ans=0x3f3f3f; 17 for(i=0;i<l;i++) 18 { 19 if(a[s[i]-‘a‘]!=-1) 20 ans=min(ans,i-a[s[i]-‘a‘]); 21 a[s[i]-‘a‘]=i; 22 } 23 if(ans==0x3f3f3f) 24 ans=-1; 25 printf("Case #%d: %d\n",T++,ans); 26 } 27 return 0; 28 }
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