http://acm.hdu.edu.cn/showproblem.php?pid=2978

题意 给出一些串 用yes表示背过了该串，问最难背的子串（子串在yes中/子串在所有字串中的个数 最小）若子串在所有字串中的个数小于m忽略
思路 暴力用map存下所有长度为k的子串及个数，最后遍历判断即可

#include <iostream>
#include <set>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
using namespace std;

const int maxn=10005;
const double eps=1e-9;
map<string,int>M;

struct Node
{
    int yes,num;
    string ss;
} a[maxn];
int main()
{
    int n,m,k;
    int flag;
    int cas=0;
    string s1,s2;
    while(scanf("%d",&n)&&n)
    {
        flag=0;
        M.clear();
        int tot=0;
        scanf("%d%d",&m,&k);
        while(n--)
        {
            cin>>s1>>s2;
            map<string,int>M2;
            M2.clear();
            int len=s1.length();
            for(int i=0; i<=len-k; i++)
            {
                string tmp="";
                for(int j=0; j<k; j++)
                    tmp+=s1[i+j];
                if(!M2[tmp])
                {
                    M2[tmp]=1;
                    int id;
                    if(M[tmp]==0)
                    {
                        M[tmp]=++tot;
                        id=tot;
                        a[id].num=0;
                        a[id].yes=0;
                        a[id].ss=tmp;
                    }
                    else
                        id=M[tmp];
                    a[id].num++;
                    if(s2=="Yes")
                        a[id].yes++;
                }
            }
            double minn=1e9;
            for(int i=1; i<=tot; i++)
            {
                if(a[i].num>=m)
                {
                    double yes1=(double)a[i].yes,num1=(double)a[i].num;
                    if(fabs(yes1/num1-minn)<eps)
                    {
                        if(a[i].num>a[flag].num)
                        {
                            flag=i;
                        }
                        else if(a[i].num==a[flag].num)
                        {
                            if(a[i].ss<a[flag].ss)
                            flag=i;
                        }
                    }
                    else if(yes1/num1<minn)
                    {
                        minn=(double)yes1/num1;
                        flag=i;
                    }
                }
            }
    }
// for(int i=1; i<=tot; i++)
// {
// cout<<a[i].ss<<" "<<a[i].yes<<" "<<a[i].num<<endl;
// }
        printf("Case %d: ",++cas);
        if(flag==0)cout<<"No solution"<<endl;
        else cout<<a[flag].ss<<endl;

    }
    return 0;
}
/* 5 2 3 EEECEGBFCBFC Yes BFCG No DEBFCEGEEC No CEG No CEG No EEE 1 1 EEC 1 2 ECE 1 1 CEG 1 4 EGB 1 1 GBF 1 1 BFC 1 3 FCB 1 1 CBF 1 1 FCE 0 2 DEB 0 1 EBF 0 1 EGE 0 1 GEE 0 1 BFC */