Codeforces 1303 C Perfect Keyboard （贪心）

Description：

Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all $26$

Polycarp uses the same password s on all websites where he is registered (it is bad, but he doesn’t care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn’t like to move his fingers while typing the password, so, for each pair of adjacent characters in s, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi… is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in $s$, so, for example, the password cannot be password (two characters $s$

Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?

Input

The first line contains one integer $T (1≤T≤1000)$

Then T lines follow, each containing one string $s (1≤|s|≤200)$

Output

For each test case, do the following:

if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);
otherwise, print YES (in upper case), and then a string consisting of $26$

Example

input

5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza

output

YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO

题意:

给你一个字符串s，现在要求你构造一个26个小写字母的一个排列t，使得给出的字符串s中，相邻字母在你构造的排列t中也是相邻的 有就输出YES 并输出构造的排列 没有就输出NO。
贪心的思路，再开一个大的数组，从中间位置开始模拟，遇到新字符看看左右两边哪里可以放，
中间进行判断。
当前字符出现过，并且上一个键入字符左右都是这个字符，肯定就输出"NO"。
没出现过，但是没有位置放也是输出"NO"。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
  int ret = 0, sgn = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9')
  {
    if (ch == '-')
      sgn = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
  {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
  if (a > 9)
    Out(a / 10);
  putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
  return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
  return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
  if (a >= mod)
    a = a % mod + mod;
  ll ans = 1;
  while (b)
  {
    if (b & 1)
    {
      ans = ans * a;
      if (ans >= mod)
        ans = ans % mod + mod;
    }
    a *= a;
    if (a >= mod)
      a = a % mod + mod;
    b >>= 1;
  }
  return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
  return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
  if (b == 0)
  {
    x = 1;
    y = 0;
    return a;
  }
  int g = exgcd(b, a % b, x, y);
  int t = x;
  x = y;
  y = t - a / b * y;
  return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
  int d, x, y;
  d = exgcd(a, p, x, y);
  if (d == 1)
    return (x % p + p) % p;
  else
    return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
  int M = 1, y, x = 0;
  for (int i = 0; i < n; ++i) //算出它们累乘的结果
    M *= a[i];
  for (int i = 0; i < n; ++i)
  {
    int w = M / a[i];
    int tx = 0;
    int t = exgcd(w, a[i], tx, y); //计算逆元
    x = (x + w * (b[i] / t) * x) % M;
  }
  return (x + M) % M;
}

const int N = 1e3 + 10;
char s[N];
char a[N];
bool vis[50];
int flag;
int l, r, pos, len;

void init(int n)
{
  mem(vis, 0);
  rep(i, 0, n << 2)
    a[i] = 0;
}

int main()
{
  int t;
  sd(t);
  while (t--)
  {
    cin >> s;
    len = strlen(s);
    pos = len * 2;
    l = 10000, r = -1;
    flag = 0;
    init(len);
    rep(i, 0, len - 1)
    {
      int x = s[i] - 'a';
      if (a[pos] == 0)
      {
        a[pos] = s[i];
      }
      else
      {
        if (a[pos] == s[i])
          continue;
        else if (a[pos + 1] != s[i] && a[pos - 1] != s[i] && vis[x])
          flag = 1;
        else if (a[pos + 1] == s[i])
          pos++;
        else if (a[pos - 1] == s[i])
          pos--;
        else if (a[pos + 1] == 0)
          a[++pos] = s[i];
        else if (a[pos - 1] == 0)
          a[--pos] = s[i];
        else
          flag = 1; //注意这一步，没出现过，但是没有位置放也是输出"NO"。
      }
      vis[x] = 1;
    }
    if (flag)
    {
      puts("NO");
      continue;
    }
    puts("YES");
    rep(i, 0, len << 2)
    {
      if (a[i] != 0)
      {
        l = min(l, i);
        r = max(r, i);
      }
    }
    rep(i, l, r)
    {
      cout << a[i];//因为按照上面的操作a数组最多放不会超过26个，所以可以先输出a数组
    }
    rep(i, 0, 25)
    {
      if (!vis[i])
        cout << (char)(i + 'a');
    }
    cout << '\n';
  }
  return 0;
}