hdu 5578 Friendship of Frog(multiset的应用)

[align=left]Problem Description[/align]

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.

[align=left]Input[/align]

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.

[align=left]Output[/align]

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.

[align=left]Sample Input[/align]

2
abcecba
abc

[align=left]Sample Output[/align]

Case #1: 2
Case #2: -1

[align=left]Source[/align]
2015ACM/ICPC亚洲区上海站-重现赛（感谢华东理工）

multiset随便搞搞就出来了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1006
#define inf 1e12
multiset<int>ms
;
multiset<int>::iterator it,it1,it2;
char s
;
int main()
{
int t;
int ac=0;
scanf("%d",&t);
while(t--){
for(int i=0;i<N;i++){
ms[i].clear();
}
scanf("%s",s);
int len=strlen(s);
for(int i=0;i<len;i++){
int u=s[i]-'a';
ms[u].insert(i);
}
int ans=1000000;
for(int i=0;i<26;i++){
//it=ms[i].begin();
int minnn=1000000;
int size_=ms[i].size();
int num=0;
for(it1=ms[i].begin();it1!=ms[i].end();it1++){
if(size_==1){
break;
}
if(num>=size_-1){
break;
}
//printf("%d= %d\n",i,(*it1));
num++;
it2=it1;

it2++;
int dis=(*it2)-(*it1);
if(dis<minnn){
minnn=dis;
}
}
//printf("minnn = %d\n",minnn);
ans=min(ans,minnn);

}
printf("Case #%d: ",++ac);
if(ans==1000000){
printf("-1\n");
}else
printf("%d\n",ans);
}
return 0;
}

View Code