hdu2987 map

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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 3EEECEGBFCBFC YesBFCG NoDEBFCEGEEC NoCEG NoCEG NoEEE 1 1EEC 1 2ECE 1 1CEG 1 4EGB 1 1GBF 1 1BFC 1 3FCB 1 1CBF 1 1FCE 0 2DEB 0 1EBF 0 1EGE 0 1GEE 0 1BFC*/

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