https://codeforces.com/contest/1481/problem/D
題意
n個點組成的完全圖，邊權爲a和b，你可以走m步，問是否有可能走出迴文串
分析
首先考慮m爲奇數的情況，如果兩點間的兩條邊相同，那麼肯定是迴文串，如果不相同，形如aba也可以構成迴文串
接着考慮m爲偶數，我們知道如果是偶數那麼中間兩個肯定是相同的，我們只要找到連着相同的兩條邊就可以。在三個點之間肯定能形成這種情況，可以自己模擬一下試試，因爲是完全圖，所以當n>=3時肯定可以構造出來的
最後再看n=2的情況，奇數條邊已經考慮過了，偶數邊要滿足必須要雙向邊都相同
Code
#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
//#define ACM_LOCAL
#define fi first
#define se second
const int N = 1e6 + 10;
const int M = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-4;
const int MOD = 1e9+7;
typedef long long ll;
typedef pair<int, int> PII;
char mp[1005][1005];
void solve() {
int T; cin >> T; while (T--) {
int n, m; cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> mp[i] + 1;
if (m & 1) {
cout << "YES" << endl;
for (int i = 1; i <= m+1; i++) cout << i % 2 + 1 << " ";
cout << endl;
continue;
}
if (n == 2) {
if (mp[1][2] == mp[2][1]) {
cout << "YES" << endl;
for (int i = 1; i <= m+1; i++) cout << i % 2 + 1 << " ";
cout << endl;
} else cout << "NO" << endl;
} else {
cout << "YES" << endl;
int flag;
if (mp[1][2] == mp[2][3]) flag = 1;
else if (mp[3][1] == mp[1][2]) flag = 0;
else flag = 2;
for (int i = 0; i <= m; i++) cout << (flag + i + m) % 3 + 1 << " ";
cout << endl;
}
}
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
solve();
}