Today is a bad day. Teemo is scolded badly by his teacher because he didn't do his homework.But Teemo is very self-confident, he tells the teacher that the problems in the homework are too simple to solve. So the teacher gets much angrier and says"I will choose a problem in the homework, if you can't solve it, I will call you mother! "
The problem is that:
There is an array A which contains n integers, and an array B which also contains n integers. You can pay one dollar to buy a card which contains two integers a1 and a2, The card can arbitrary number of times transform a single integer a1 to a2 and vise-versa on both array A and Array B. Please calculate the minimum dollars you should pay to make the two array same(For every 1<=i<=n,A[i]=B[i]);
Input Format
- The first line of the input contains an integer T(1<=T<=10), giving the number of test cases.
- For every test case, the first line contains an integer n(1<=n<=500000). The second line contains n integers. The i th integer represents A[i](1<=A[i]<=100000). And the third line contains n integers. The i th integer represents B[i](1<=B[i]<=100000).
Output Format
For each test case, output an integer which means the minimum dollars you should pay in a line.
样例输入
1 5 1 1 2 3 2 1 2 3 1 1
样例输出
2
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <string> 6 #include <string> 7 #include <map> 8 #include <cmath> 9 #include <set> 10 #include <ctime> 11 #include <algorithm> 12 using namespace std; 13 const int N=5e5+9; 14 const int M=1e5+9; 15 using namespace std; 16 int t,n,a[N],b[N]; 17 //map<int,int>mp; 18 int mp[M]; 19 vector<int>ve[M]; 20 int ans1,ans2; 21 bool vis[M]; 22 void dfs(int x) 23 { 24 vis[x]=1; 25 for(int i=0;i<ve[x].size();i++){ 26 int y=ve[x][i]; 27 if(!vis[y]) 28 { 29 dfs(y); 30 } 31 } 32 } 33 int main() 34 { 35 scanf("%d",&t); 36 while(t--) 37 { clock_t sta=clock(); 38 scanf("%d",&n); 39 //mp.erase(mp.begin(),mp.end());//用map会超时 40 memset(mp,0,sizeof(mp)); 41 //memset(vis,0,sizeof(vis)); 42 for(int i=1;i<=M;i++) ve[i].clear(); 43 for(int i=1;i<=n;i++) 44 { 45 scanf("%d",&a[i]); 46 } 47 for(int i=1;i<=n;i++) scanf("%d",&b[i]); 48 for(int i=1;i<=n;i++) { 49 if(a[i]!=b[i]){ 50 mp[a[i]]=1; 51 mp[b[i]]=1; 52 ve[a[i]].push_back(b[i]); 53 ve[b[i]].push_back(a[i]); 54 } 55 } 56 ans1=ans2=0; 57 memset(vis,0,sizeof(vis)); 58 //将该子联通快的所有点和根相互交换 59 for(int i=1;i<=M;i++){ 60 if(mp[i]==1){ 61 ans1++; 62 if(vis[i]==0){ 63 dfs(i); 64 ans2++; 65 } 66 } 67 } 68 printf("%d\n",ans1-ans2); 69 clock_t end=clock(); 70 //printf("%d\n",end-sta); 71 cout<<end-sta<<endl; 72 } 73 return 0; 74 }