構造 ---- D. AB Graph(偶數和3遠環的關系)

題目連結

題目大意：

n n n個點組成的有向完全圖，邊權為 a a a和 b b b，你可以走 m m m步，問是否有可能走出回文串??

解題思路：

1. 首先我們發現如果 ( u , v ) = ( v , u ) (u,v)=(v,u) (u,v)=(v,u)的話就可以來回走了
2. 如果 m m m是奇數那麼在一條邊裡面來回就可以構造了
3. 如果 m m m是偶數：我們要找到一個三元回路，回路上面有兩個相同的字母！
4. 因為字母隻有 a , b a,b a,b那麼很明顯任意一個三元回路都可以（鴿巢原理）
5. 我們觀察一下如果全是 a a a aaa aaa那麼肯定沒什麼好讨論的
6. 環可以是 a b a aba aba,那麼我們看看 a b a    a b a    a b a    a b a . . . . aba\;aba\;aba\;aba .... abaabaabaaba....
7. 找規律我們可以發現從 m % 3 m\%3 m%3的位置開始取 m m m位一定是回文串

ACcode

#include <bits/stdc++.h>
#define mid ((l + r) >> 1) 
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define hash Hash
#define next Next
#define f first
#define s second
using namespace std;
const int maxn = 1e5 + 10;
const double eps = 1e-9;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
int _;
int n, k;
char ch[1005][1005];

void slove() {
    int center = -1;
    PII taga = {-1,-1};
    for(int i = 1; i <= n; ++ i) {
       for(int j = 1; j <= n; ++ j) {
           if(ch[i][j] == 'a') taga.first = j;
           else if(ch[i][j] == 'b') taga.second = j;
       }
       if(taga.first != -1 && taga.second != -1) {
           center = i;
           break;
       }
       taga = {-1,-1};
    }
   cout << "YES\n";
   int a[] = {center,taga.first,taga.second};
   int b[] = {taga.first,taga.second,center};
   int c[] = {taga.second,center,taga.first};
   for(int i = 0; i <= k; ++ i) {
       if(k % 3 == 0) cout << a[i%3] << " ";
       else if(k % 3 == 1) cout << b[i%3] << " ";
       else cout << c[i%3] << " ";
   }
   cout << endl;
   return;
}

int main()
{
    IOS;
    cin >> _;
    while(_--) {
        cin >> n >> k;
        for(int i = 1; i <= n; ++ i) 
            cin >> (ch[i] + 1);
        int u, v;
        for(int i = 1; i <= n ; ++ i)
          for(int j = 1; j <= n ; ++ j) {
              if(i == j) continue;
              if(ch[i][j] == ch[j][i]) {
                  u = i, v = j;
                  goto loop;
              }
          }
        
        if(k & 1) {
            cout << "YES\n";
            for(int i = 1; i <= k + 1; ++ i)
               if(i & 1) cout << "1 ";
               else cout << "2 ";
             cout << endl;
            
        } else {
            if(n == 2) cout << "NO\n";
            else slove();
        }
        
        continue;
        
        loop:
        // 有對稱的邊來回走就可以了
        cout << "YES\n";
        for(int i = 1; i <= k + 1; ++ i)
           if(i & 1) cout << u << " ";
           else cout << v << " ";
        cout << endl;
    }

}