hdu 5578 Friendship of Frog 2015ACM/ICPC亚洲区上海站-重现赛

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将同一个country的frog放在一起，按照index从小到大排序。因此minimum distance一定出现在相邻的两个frog中。

#include<iostream>#include<stdio.h>#include<cstdio>#include<string>#include<cmath>#include<stdlib.h>#include<algorithm>#include<string.h>#include<cstring>#include<vector>#include<queue>#include<map>using namespace std;//hdu 5578const int maxn=1010;const int INF=0x3f3f3f3f;int T;int N;int ans;class frog{public:    int country;    int idx;public:    frog()    {        country=0;        idx=0;    }    frog(int c,int id)    {        country=c;        idx=id;    }    bool operator <(const frog a) const    {        return idx<a.idx;    }};vector<frog>mp[27];int main(){    freopen("input.txt","r",stdin);    scanf("%d",&T);    for(int ca=1;ca<=T;ca++)    {        char str[maxn];        memset(mp,0,sizeof(mp));        scanf("%s",&str);        for(int i=0;i<strlen(str);i++)        {            mp[str[i]-'a'].push_back(frog(str[i]-'a',i+1));        }        ans=INF;        for(int i=0;i<26;i++)        {            sort(mp[i].begin(),mp[i].end());            for(int j=1;j<mp[i].size();j++)            {                ans=min(ans,mp[i][j].idx-mp[i][j-1].idx);            }        }        if(ans==INF) ans=-1;        printf("Case #%d: %d\n",ca,ans);    }}

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