ACM训练联盟周赛 A. Teemo's bad day

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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

样例输出

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <vector>

 #include <string>

 #include <string>

 #include <map>

 #include <cmath>

 #include <set>

 #include <ctime>

 #include <algorithm>

 using namespace std;

 const int N=5e5+;

 const int M=1e5+;

 using namespace std;

 int t,n,a[N],b[N];

 //map<int,int>mp;

 int mp[M];

 vector<int>ve[M];

 int ans1,ans2;

 bool vis[M];

 void dfs(int x)

 {

     vis[x]=;

     for(int i=;i<ve[x].size();i++){

         int y=ve[x][i];

         if(!vis[y])

         {

             dfs(y);

         }

     }

 }

 int  main()

 {

     scanf("%d",&t);

     while(t--)

     {    clock_t sta=clock();

         scanf("%d",&n);

         //mp.erase(mp.begin(),mp.end());//用map会超时

         memset(mp,,sizeof(mp));

         //memset(vis,0,sizeof(vis));

         for(int i=;i<=M;i++)  ve[i].clear();

         for(int i=;i<=n;i++)

         {

             scanf("%d",&a[i]);

         }

         for(int i=;i<=n;i++)  scanf("%d",&b[i]);

         for(int i=;i<=n;i++)  {

             if(a[i]!=b[i]){

                 mp[a[i]]=;

                 mp[b[i]]=;

                 ve[a[i]].push_back(b[i]);

                 ve[b[i]].push_back(a[i]);//一定是无向图，不然可能一个联通快走不遍

             }

         }

         ans1=ans2=;

         memset(vis,,sizeof(vis));

         //将该子联通快的所有点和根相互交换

         for(int i=;i<=M;i++){

            if(mp[i]==){

                ans1++;

                if(vis[i]==){

                    dfs(i);

                    ans2++;

                }

            }

         }

         printf("%d\n",ans1-ans2);

         clock_t end=clock();

         //printf("%d\n",end-sta);

         cout<<end-sta<<endl;

     }

     return ;

 }