1063 Set Similarity (25)
Given two sets of integers, the similarity of the sets is defined to be Nc/Nt×100%, where Nc is the number of distinct common numbers shared by the two sets, and Nt 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 (≤10^4) and followed by M integers in the range [0,10^9]. 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:
3
3 99 87 101
4 87 101 5 87
7 99 101 18 5 135 18 99
2
1 2
1 3
Sample Output:
50.0%
33.3%题目大意：给定两个整数集合，它们的相似度定义为：Nc/Nt*100%。其中Nc是两个集合都有的不相等整数的个数，Nt是两个集合一共有的不相等整数的个数。你的任务就是计算任意一对给定集合的相似度。nc是两个集合的公共元素个数，nt是两个集合的所有包含的元素个数（其中元素个数表示各个元素之间互不相同）
分析：题目要求为求两个数组交集中的元素数量Nc，两个数组并集中的元素数量Nt，再求这两个数的商。注意同一个数组中的重复元素只计入一次。除了模拟，还可用set，或者哈希。
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#define ll long long
#define INF 0x3f3f3f3f
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
using namespace std;bool cmp(int a,int b)
{return a<b;
}int main(void)
{#ifdef testfreopen("in.txt","r",stdin);//freopen("in.txt","w",stdout);clock_t start=clock();#endif //testint n;scanf("%d",&n);int num[n+5][10010]={0},cnt[n+5]={0},temp[n+5][10010]={0};for(int i=1;i<=n;++i){scanf("%d",&cnt[i]);for(int j=0;j<cnt[i];++j)scanf("%d",&temp[i][j]);}for(int i=1;i<=n;++i)sort(temp[i],temp[i]+cnt[i]);for(int i=1;i<=n;++i){int t=0;num[i][0]=temp[i][0];for(int j=1;j<cnt[i];++j)if(temp[i][j]!=temp[i][j-1])t++,num[i][t]=temp[i][j];cnt[i]=t+1;}int k;scanf("%d",&k);for(int ii=0;ii<k;++ii){int a,b,nc=0,nt=0;scanf("%d%d",&a,&b);int l1=0,l2=0;//db2(cnt[a],cnt[b]);while(l1<cnt[a]||l2<cnt[b]){nt++;if(l1<cnt[a]&&l2<cnt[b]){if(num[a][l1]>num[b][l2])l2++;else if(num[a][l1]<num[b][l2])l1++;else nc++,l1++,l2++;}else if(l1<cnt[a])l1++;else l2++;}//db4(nc,nt,l1,l2);printf("%.1f%\n",nc*100.0/nt);}#ifdef testclockid_t end=clock();double endtime=(double)(end-start)/CLOCKS_PER_SEC;printf("\n\n\n\n\n");cout<<"Total time:"<<endtime<<"s"<<endl; //s为单位cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms为单位#endif //testreturn 0;
}