PAT 1063. Set Similarity (25) 寻找交集。未解之谜。

1063. Set Similarity (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given two sets of integers, the similarity of the sets is defined to be N_c/N_t*100%, where N_c is the number of distinct common numbers shared by the two sets, and N_t is the total number of distinct numbers in the two sets. Your job is to calculate the similarity of any given pair of sets.

Input Specification:

Each input file contains one test case. Each case first gives a positive integer N (<=50) which is the total number of sets. Then N lines follow, each gives a set with a positive M (<=10⁴) and followed by M integers in the range [0, 10⁹]. After the input of sets, a positive integer K (<=2000) is given, followed by K lines of queries. Each query gives a pair of set numbers (the sets are numbered from 1 to N). All the numbers in a line are separated by a space.

Output Specification:

For each query, print in one line the similarity of the sets, in the percentage form accurate up to 1 decimal place.

Sample Input:

33 99 87 1014 87 101 5 877 99 101 18 5 135 18 9921 21 3

Sample Output:

50.0%
33.3%
不懂为啥把那两行注释掉就过了。感觉应该没影响的。。。
#include<string>#include<stdio.h>#include<iostream>#include<string.h>#include<queue>#include<algorithm>#include<map>#include<set>#include<math.h>#include<stack>#include<vector>using namespace std;set<int> node[52];void judge(int x,int y){    set<int>::iterator it1=node[x].begin();    set<int>::iterator it2=node[y].begin();    int last=-1;    int sum1=node[x].size()+node[y].size();    int sum2=0;    while(it1!=node[x].end()&&it2!=node[y].end())      {          if(*it1==*it2)          {              sum2++;              //last=-1;              it1++;              it2++;          }          else if(*it1<*it2)          {           //  if(*it1==last) sum2++;    这两行一定要注释掉！！             // last=*it1;              it1++;          }          else {            //  if(*it2==last) sum2++;    这两行一定要注释掉！！             // last=*it2;              it2++;                }     }  float tmp=(float)sum2/(float)(sum1-sum2);             int zh=int(tmp*1000+0.5);             int xx,yy,zz;             xx=zh/100;             yy=(zh%100)/10;             zz=zh%10;            if(xx!=0) cout<<xx;            cout<<yy<<'.'<<zz<<'%'<<endl;}int main(){     int n,k;     cin>>n;    for(int i=1;i<=n;i++)    {        int tmp;        scanf("%d",&tmp);        for(int j=0;j<tmp;j++)        {            int tmmp;            scanf("%d",&tmmp);            node[i].insert(tmmp);        }    }    cin>>k;    for(int i=0;i<k;i++)    {        int a1,b1;        scanf("%d %d",&a1,&b1);        judge(a1,b1);    }     return 0;}