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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 the2nd 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.
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.
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.
2 abcecba abc
Case #1: 2 Case #2: -1
multiset随便搞搞就出来了。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 #include <stack> 15 using namespace std; 16 #define PI acos(-1.0) 17 #define max(a,b) (a) > (b) ? (a) : (b) 18 #define min(a,b) (a) < (b) ? (a) : (b) 19 #define ll long long 20 #define eps 1e-10 21 #define MOD 1000000007 22 #define N 1006 23 #define inf 1e12 24 multiset<int>ms[N]; 25 multiset<int>::iterator it,it1,it2; 26 char s[N]; 27 int main() 28 { 29 int t; 30 int ac=0; 31 scanf("%d",&t); 32 while(t--){ 33 for(int i=0;i<N;i++){ 34 ms[i].clear(); 35 } 36 scanf("%s",s); 37 int len=strlen(s); 38 for(int i=0;i<len;i++){ 39 int u=s[i]-‘a‘; 40 ms[u].insert(i); 41 } 42 int ans=1000000; 43 for(int i=0;i<26;i++){ 44 //it=ms[i].begin(); 45 int minnn=1000000; 46 int size_=ms[i].size(); 47 int num=0; 48 for(it1=ms[i].begin();it1!=ms[i].end();it1++){ 49 if(size_==1){ 50 break; 51 } 52 if(num>=size_-1){ 53 break; 54 } 55 //printf("%d= %d\n",i,(*it1)); 56 num++; 57 it2=it1; 58 59 it2++; 60 int dis=(*it2)-(*it1); 61 if(dis<minnn){ 62 minnn=dis; 63 } 64 } 65 //printf("minnn = %d\n",minnn); 66 ans=min(ans,minnn); 67 68 } 69 printf("Case #%d: ",++ac); 70 if(ans==1000000){ 71 printf("-1\n"); 72 }else 73 printf("%d\n",ans); 74 } 75 return 0; 76 }
hdu 5578 Friendship of Frog(multiset的应用)
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原文地址：http://www.cnblogs.com/UniqueColor/p/5011296.html
