CF - 791C. Bear and Different Names - 贪心+模拟

1.题目描述：

C. Bear and Different Names

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

In the army, it isn't easy to form a group of soldiers that will be effective on the battlefield. The communication is crucial and thus no two soldiers should share a name (what would happen if they got an order that Bob is a scouter, if there are two Bobs?).

A group of soldiers is effective if and only if their names are different. For example, a group (John, Bob, Limak) would be effective, while groups (Gary, Bob, Gary) and (Alice, Alice) wouldn't.

You are a spy in the enemy's camp. You noticed n soldiers standing in a row, numbered 1 through n.
The general wants to choose a group of k consecutive soldiers. For every k consecutive
soldiers, the general wrote down whether they would be an effective group or not.

You managed to steal the general's notes, with n - k + 1 strings s1, s2, ..., sn - k + 1,
each either "YES" or "NO".

The string s1 describes
a group of soldiers 1 through k ("YES"
if the group is effective, and "NO" otherwise).

The string s2 describes
a group of soldiers 2 through k + 1.

And so on, till the string sn - k + 1 that
describes a group of soldiers n - k + 1 through n.

Your task is to find possible names of n soldiers. Names should match the stolen notes. Each name should be a string that consists
of between 1 and 10 English
letters, inclusive. The first letter should be uppercase, and all other letters should be lowercase. Names don't have to be existing names — it's allowed to print "Xyzzzdj"
or "T" for example.

Find and print any solution. It can be proved that there always exists at least one solution.

Input

The first line of the input contains two integers n and k (2 ≤ k ≤ n ≤ 50) —
the number of soldiers and the size of a group respectively.

The second line contains n - k + 1 strings s1, s2, ..., sn - k + 1.
The string si is
"YES" if the group of soldiers i through i + k - 1 is
effective, and "NO" otherwise.

Output

Find any solution satisfying all given conditions. In one line print n space-separated strings, denoting possible names of soldiers
in the order. The first letter of each name should be uppercase, while the other letters should be lowercase. Each name should contain English letters only and has length from 1 to 10.

If there are multiple valid solutions, print any of them.

Examples

input

8 3
NO NO YES YES YES NO

output

Adam Bob Bob Cpqepqwer Limak Adam Bob Adam

input

9 8
YES NO

output

R Q Ccccccccc Ccocc Ccc So Strong Samples Ccc

input

3 2
NO NO

output

Na Na Na

Note

In the first sample, there are 8 soldiers. For every 3 consecutive
ones we know whether they would be an effective group. Let's analyze the provided sample output:

First three soldiers (i.e. Adam, Bob, Bob) wouldn't be an effective group because there are two Bobs. Indeed, the string s1 is
"NO".

Soldiers 2 through 4 (Bob,
Bob, Cpqepqwer) wouldn't be effective either, and the string s
eb45
2 is
"NO".

Soldiers 3 through 5 (Bob,
Cpqepqwer, Limak) would be effective, and the string s3 is
"YES".

...,

Soldiers 6 through 8 (Adam,
Bob, Adam) wouldn't be effective, and the string s6 is
"NO".

2.题意概述：

有n个士兵站成一排，他们的名字分别为S1、S2......Sn，从1 到 n - k + 1分别给出从i开始连续k个士兵组成的小队伍中是否有名字相同的判断，要你构造出这样一支队伍满足所给的条件。

3.解题思路：

既然是构造，那就模拟呀。从1开始直到第一个YES的位置开始，贪心地给这k个士兵分别起不同的名字（ 这样位置为0~i - 1个no的士兵都起相同的名字。因为包括他们的话，就是有重名嘛）。那么再 令j从i + 1开始往后推到n-k+1，如果第j个位置是YES代表从j到j+k个士兵又满足名字不相同，那么就给它（第j + k个士兵）赋一个新的名字，如果是NO，那就贪心地给它第j个名字（这样可以保证第j 个士兵和第j + k个士兵名字冲突，且不会影响到前j次做出的选择）。还有就是如果26个大写字母不够用就再加小写字母，但是要保证第一个字母一定是大写（这里我wa了一发，没认真读题！！！）
这里其实最多只要两个长度字符就足够，因为26^2 > 50，肯定可以完全区分地表示出所有士兵名字。

4.AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define N 51
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
string s
;
int ans
;
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
long _begin_time = clock();
#endif
int n, k;
while (~scanf("%d%d", &n, &k))
{
memset(ans, 0, sizeof(ans));
for (int i = 0; i < n - k + 1; i++)
cin >> s[i];
int i = 0;
while (i < n - k + 1 && s[i][0] == 'N')
i++;
int cnt = 1;
for (int j = i + 1; j < i + k; j++)
{
ans[j] = ans[j - 1] + 1;
cnt++;
}
for (int j = i + 1; j < n - k + 1; j++)
if (s[j][0] == 'Y')
ans[j + k - 1] = cnt++;
else
ans[j + k - 1] = ans[j];
int first = 1;
for (i = 0; i < n; i++)
if (ans[i] + 'A' > 'Z')
{
char cur[3];
cur[0] = ans[i] + 'A' - 'Z' + 'A';
cur[1] = ans[i] + 'A' - 'Z' + 'a';
cur[2] = '\0';
if (first)
{
first = 0;
printf("%s", cur);
}
else
printf(" %s", cur);
}
else
{
if (first)
{
first = 0;
printf("%c", ans[i] + 'A');
}
else
printf(" %c", ans[i] + 'A');
}
puts("");
}

#ifndef ONLINE_JUDGE
long _end_time = clock();
printf("time = %ld ms.", _end_time - _begin_time);
#endif
return 0;
}