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

1
5
1 1 2 3 2
1 2 3 1 1

样例输出

2

#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+9;
const int M=1e5+9;
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]=1;
    for(int i=0;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,0,sizeof(mp));
        //memset(vis,0,sizeof(vis));
        for(int i=1;i<=M;i++)  ve[i].clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=n;i++)  scanf("%d",&b[i]);
        for(int i=1;i<=n;i++)  {
            if(a[i]!=b[i]){
                mp[a[i]]=1;
                mp[b[i]]=1;
                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=0;
        memset(vis,0,sizeof(vis));
        //将该子联通快的所有点和根相互交换
        for(int i=1;i<=M;i++){
           if(mp[i]==1){
               ans1++;
               if(vis[i]==0){
                   dfs(i);
                   ans2++;
               }
           }    
        }
        printf("%d
",ans1-ans2);
        clock_t end=clock();
        //printf("%d
",end-sta);
        cout<<end-sta<<endl;
    }
    return 0;
}