对于1<=i<=k, 把 a[c[i]] 改为c[i % k + 1]。给定n，k和数组b，判断能否得到数组b。

题目

思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5;
const int mod = 1e9 + 7;
// const int mod = 998244353;
const int N = 1e6;
// int a[505][5005];
// bool vis[505][505];
// char s[505][505];
int a[maxn], b[maxn];
bool vis[maxn];
string s;
int n, m;struct Node{int val, id;bool operator<(const Node &u)const{return val < u.val;}// int x, y;// int l, r, j;
}c[maxn];int ans[maxn];vector<int> G[maxn], g[maxn];
int dfn[maxn], low[maxn], in_st[maxn];
int col[maxn], cnt_col, tot;
int sum[maxn], siz[maxn];
stack<int> st;void tarjan(int u){dfn[u] = ++tot;low[u] = dfn[u];in_st[u] = 1;st.push(u);for(auto v : G[u]){if(!dfn[v]){//v没遍历过，先更新v，更新utarjan(v);low[u] = min(low[u], low[v]);}else{if(in_st[v]) //后向边或者横向边，就更新low[u] = min(low[u], dfn[v]);}}if(low[u] == dfn[u]){//找到该强连通分量的最先访问到的点col[u] = ++cnt_col;//把该分量的点都标记为同一颜色sum[cnt_col] += a[u];siz[cnt_col]++;while(!st.empty() && st.top() != u){int x = st.top();st.pop();in_st[x] = 0;col[x] = cnt_col;sum[cnt_col] += a[x];siz[cnt_col]++;}if(!st.empty()){st.pop();in_st[u] = 0;}}
}void solve(){int res = 0;int q, k;cin >> n >> k;tot = 0;cnt_col = 0;for(int i = 1; i <= n; i++){G[i].clear();g[i].clear();siz[i] = 0;in_st[i] = 0;dfn[i] = low[i] = 0;}while(!st.empty()){st.pop();}for(int i = 1; i <= n; i++){cin >> a[i];G[i].pb(a[i]);}if(k == 1){for(int i = 1; i <= n; i++){if(a[i] != i){cout << "No\n";return;}}cout << "Yes\n";return;}for(int i = 1; i <= n; i++){if(a[i] == i){cout << "No\n";//不能自环return;}if(!dfn[i]){tarjan(i);}}for(int i = 1; i <= cnt_col; i++){if(siz[i] > 1 && siz[i] != k){cout << "No\n";return;}}cout << "Yes\n";}signed main(){ios::sync_with_stdio(0);cin.tie(0);int T = 1;cin >> T;while (T--){solve();}return 0;
}

法二：

#include <bits/stdc++.h>using i64 = long long;void solve() {int n, k;std::cin >> n >> k;std::vector<int> b(n);for (int i = 0; i < n; i++) {std::cin >> b[i];b[i]--;}if (k == 1) {for (int i = 0; i < n; i++) {if (b[i] != i) {std::cout << "NO\n";return;}}std::cout << "YES\n";return;}std::vector<int> vis(n, -1);for (int i = 0; i < n; i++) {int j = i;while (vis[j] == -1) {vis[j] = i;j = b[j];}if (vis[j] == i) {int len = 0;int x = j;do {len++;x = b[x];} while (x != j);if (len != k) {std::cout << "NO\n";return;}}}std::cout << "YES\n";
}int main() {std::ios::sync_with_stdio(false);std::cin.tie(nullptr);int t;std::cin >> t;while (t--) {solve();}return 0;
}