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 Nlines 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:
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%首先附上代码：
#include "cstdio"
#include "cstring"
#include "set"
using namespace std;
int main(){
int n;
scanf("%d",&n);
set<int> s[n];
for (int i=0; i<n; i++) {
int m;
scanf("%d",&m);
for (int j=0; j<m; j++) {
int current;
scanf("%d",¤t);
s[i].insert(current);
}
}
int k;
scanf("%d",&k);
for (int i=0; i<k; i++) {
int x,y;
scanf("%d %d",&x,&y);
set<int>::iterator t1,t2;
double Nc=0,Nt=0;
for (t1=s[x-1].begin(),t2=s[y-1].begin(); t1!=s[x-1].end()&&t2!=s[y-1].end();) {
//printf(" -%d-%d- ",*t1,*t2);
if (*t1==*t2) { //相当则分子加1，同时后移一个继续比较
Nc++;
t1++;
t2++;
}
else if(*t1<*t2){
Nt++;
t1++;
}
else{
Nt++;
t2++;
}
}
if (t2!=s[y-1].end()) { //t2更长
while (t2!=s[y-1].end()) {
Nt++;
t2++;
}
}
else if(t1!=s[x-1].end()){
while (t1!=s[x-1].end()) { //t1更长
Nt++;
t1++;
}
}
printf("%.1f%%\n",Nc/(Nt+Nc)*100);
}}
反思：
没什么难点，主要就是set的用法，要记住set是自动升序排列的，自动去重的，添加用insert()
