CodeForces - 1237B - Balanced Tunnel(思维)

题目链接：https://vjudge.net/problem/CodeForces-1237B

题目大意：给你两个长度为n的数列，看看第二个数列有多少数在原数列数字的前面

　　从原数列的角度出发，第二个数列中和原数列相等的数前面多出来的数就是移动后的数

#include<set>
#include<map>
#include<list>
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define endl '\n'
#define rtl rt<<1
#define rtr rt<<1|1
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define maxx(a, b) (a > b ? a : b)
#define minn(a, b) (a < b ? a : b)
#define zero(a) memset(a, 0, sizeof(a))
#define INF(a) memset(a, 0x3f, sizeof(a))
#define IOS ios::sync_with_stdio(false)
#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<char, int> P2;
const double pi = acos(-1.0);
const double eps = 1e-7;
const ll MOD =  1000000007LL;
const int INF = 0x3f3f3f3f;
const int _NAN = -0x3f3f3f3f;
const double EULC = 0.5772156649015328;
const int NIL = -1;
template<typename T> void read(T &x){
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
const int maxn = 1e5+10;
int a[maxn], b[maxn], vis[maxn];
int main(void) {
    int n;
    scanf("%d", &n);
    for (int i = 0; i<n; ++i)
        scanf("%d", &a[i]);
    for (int i = 0; i<n; ++i)
        scanf("%d", &b[i]);
    int l = 0, ans = 0;
    for (int i = 0; i<n && l<n-1; ++i) { //对第二数列的最后一个数比较是没有意义的，所以n-1
        if (vis[a[i]]) continue;
        while(l<n-1 && a[i]!=b[l]) {
            vis[b[l++]] = true; //对移出去的数字做标记
            ++ans;
        }
        b[l++] = true; //此时两数列的两个数相等，移动到下一个位置进行比较
    }
    printf("%d\n", ans);
    return 0;
}