HDU 5495（置换群)

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两个数组a和b，将其看做一个由a到b的置换，那么可以得到若干个环，显然环之间是独立的。不难看出对每个长度大于1的环，一定可以得到一个len-1的LCS。所以答案就是n-长度大于1的循环的个数。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define rep(i, n) for(int i = 0; i < (n); i ++)#define REP(i, t, n) for(int i = (t); i < (n); i ++)#define FOR(i, t, n) for(int i = (t); i <= (n); i ++)#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_backconst int inf = 0x3f3f3f3f, N = 1e5 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;int a[N], b[N], vis[N], hs[20 * N];// Impint main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    si(T);    while(T --) {    mem(vis, 0);    si(n);    rep(i, n) si(a[i]), hs[a[i]] = i;    rep(i, n) si(b[i]);    int num = 0;    rep(i, n) {    if(!vis[i]) {    int id = i, cnt = 0;    do {    vis[id] = 1;    cnt ++;    id = hs[b[id]];    } while(id != i);    num += cnt > 1;    }    }    printf("%d\n", n - num);    }        return 0;}

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