L2-005. 集合相似度

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开始有一题一直超时，然后不断想着改进的办法。

然后想到一种，既然每次都要寻找一个f1值其包含的值，那么可以先一次找完所有这个f1有的数。

set对于数字的查找速度远远大于map！！！map是比较慢的

#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>#include <map>#include <set>using namespace std;struct ttt {int a[1050];int s;};ttt qq[55];double walk[55][55];set<int>vis1[55];int summ[55];bool walked[55][55];void init(int n){    for(int i=1;i<=qq[n].s;i++){        if(vis1[n].count(qq[n].a[i])==0){                vis1[n].insert(qq[n].a[i]);            summ[n]++;        }    }    //cout << summ[n] << endl;}double shou(int f1,int f2){    set<int>vis2;    set<int>vis3;    int i,j,k;    double sum1,sum2;    sum1=summ[f1];    sum2=0;   // cout << "777" << endl;    for(i=1;i<=qq[f2].s;i++){     //       cout << vis1.count(qq[f2].a[i]) <<endl;   // cout << vis2.count(qq[f2].a[i]) <<endl;    //cout << endl;        if(vis1[f1].count(qq[f2].a[i])==0&&vis3.count(qq[f2].a[i])==0){           vis3.insert(qq[f2].a[i]);           sum1++;        }else if(vis1[f1].count(qq[f2].a[i])==1&&vis2.count(qq[f2].a[i])==0){            vis2.insert(qq[f2].a[i]);           sum2++;        }    }     //   cout << sum1 << "    "<< sum2 << endl;    return sum2/sum1;}int main(){    //freopen("in.txt","r",stdin);    int i,j,k,l,f2,f1,f3,t1,t2,t3;    int n,m;    memset(summ,0,sizeof(summ));    cin >>n;    for(i=1;i<=n;i++){        scanf("%d",&qq[i].s);        for(j=1;j<=qq[i].s;j++)            scanf("%d",&qq[i].a[j]);    }    cin >> m;    double sum1;    memset(walk,0,sizeof(walk));    for(i=1;i<=n;i++){        init(i);    }    /*for(i=1;i<n;i++)    for(j=i+1;j<=n;j++){        sum1=shou(i,j);        walk[i][j]=sum1;    }*/    memset(walked,0,sizeof(walked));    for(i=1;i<=m;i++){        scanf("%d%d",&f1,&f2);       // sum1=shou(f1,f2);       if(f2<f1)swap(f1,f2);        if(walked[f1][f2]==0){          walk[f1][f2]=shou(f1,f2);          walked[f1][f2]=1;        }        printf("%.2f",walk[f1][f2]*100);        cout << "%" <<endl;    }    return 0;}

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