UVaLive 3641 Leonardo's Notebook (置换)

题意：给定一个置换 B 问是否则存在一个置换 A ，使用 A^2 = B。

析：可以自己画一画，假设 A = (a1, a2, a3)(b1, b2, b3, b4)，那么 A^2 = (a1, a2, a3)(b1, b2, b3, b4)(a1, a2, a3)(b1, b2, b3, b4)，不相关循环可以有交换律。

A^2 = (a1, a2, a3)(a1, a2, a3)(b1, b2, b3, b4)(b1, b2, b3, b4)，分别考虑这两个循环，可以得到两个奇循环置换后仍然是一个奇循环，而两个偶循环置换后就是两个 n/2 循环，n 是 循环的长度，所以对于这个题目，B 中的所有奇循环，都可以从 A 得到，而偶循环，必须两两配对，才能存在，配对指的是长度一样。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 30 + 10;
const int maxm = 1e6 + 2;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

char s[maxn];
bool vis[maxn];

int main(){
  int T;  cin >> T;  n = 26;
  while(T--){
    scanf("%s", s);  ms(vis, 0);
    vector<int> v;
    for(int i = 0; i < n; ++i)  if(!vis[i]){
      int cnt = 1, x = i;
      while(s[x] != i + 'A'){
        ++cnt;  x = s[x] - 'A';
        vis[x] = 1;
      }
      v.pb(cnt);
    }
    sort(v.begin(), v.end());  v.pb(-1);
    bool ok = true;
    for(int i = 0; i < v.sz && ok; ++i)  
      if(v[i] % 2 == 0){
        if(v[i] != v[i+1])  ok = false;
        else ++i;
      }
    puts(ok ? "Yes" : "No");
  }
  return 0;
}