CF1394C-Boboniu and String
题目:
题目描述:
Boboniu defines BN-string as a string $ s $ of characters ‘B’ and ‘N’.
You can perform the following operations on the BN-string $ s $ :
- Remove a character of $ s $ .
- Remove a substring “BN” or “NB” of $ s $ .
- Add a character ‘B’ or ‘N’ to the end of $ s $ .
- Add a string “BN” or “NB” to the end of $ s $ .
Note that a string $ a $ is a substring of a string $ b $ if $ a $ can be obtained from $ b $ by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Boboniu thinks that BN-strings $ s $ and $ t $ are similar if and only if:
- $ |s|=|t| $ .
- There exists a permutation $ p_1, p_2, \ldots, p_{|s|} $ such that for all $ i $ ( $ 1\le i\le |s| $ ), $ s_{p_i}=t_i $ .
Boboniu also defines $ \text{dist}(s,t) $ , the distance between $ s $ and $ t $ , as the minimum number of operations that makes $ s $ similar to $ t $ .
Now Boboniu gives you $ n $ non-empty BN-strings $ s_1,s_2,\ldots, s_n $ and asks you to find a non-empty BN-string $ t $ such that the maximum distance to string $ s $ is minimized, i.e. you need to minimize $ \max_{i=1}^n \text{dist}(s_i,t) $ .
输入格式:
The first line contains a single integer $ n $ ( $ 1\le n\le 3\cdot 10^5 $ ).
Each of the next $ n $ lines contains a string $ s_i $ ( $ 1\le |s_i| \le 5\cdot 10^5 $ ). It is guaranteed that $ s_i $ only contains ‘B’ and ‘N’. The sum of $ |s_i| $ does not exceed $ 5\cdot 10^5 $ .
输出格式:
In the first line, print the minimum $ \max_{i=1}^n \text{dist}(s_i,t) $ .
In the second line, print the suitable $ t $ .
If there are several possible $ t $ ’s, you can print any.
样例:
样例输入 1:
样例输出 1:
样例输入 2:
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| 10
N
BBBBBB
BNNNBBNBB
NNNNBNBNNBNNNBBN
NBNBN
NNNNNN
BNBNBNBBBBNNNNBBBBNNBBNBNBBNBBBBBBBB
NNNNBN
NBBBBBBBB
NNNNNN
|
样例输出 2:
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| 12
BBBBBBBBBBBBNNNNNNNNNNNN
|
样例输入 3:
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9
| 8
NNN
NNN
BBNNBBBN
NNNBNN
B
NNN
NNNNBNN
NNNNNNNNNNNNNNNBNNNNNNNBNB
|
样例输出 3:
样例输入 4:
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| 3
BNNNBNNNNBNBBNNNBBNNNNBBBBNNBBBBBBNBBBBBNBBBNNBBBNBNBBBN
BBBNBBBBNNNNNBBNBBBNNNBB
BBBBBBBBBBBBBBNBBBBBNBBBBBNBBBBNB
|
样例输出 4:
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2
| 12
BBBBBBBBBBBBBBBBBBBBBBBBBBNNNNNNNNNNNN
|
思路:
实现:
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| #include <bits/stdc++.h>
using namespace std;
#ifndef use_ios11
#define use_ios11
using namespace std;
struct ins
{
int ans;
ins() { ans = 1; }
#define endl '\n'
void read()
{
}
void read1(char &s)
{
char c = getchar();
for (; !isprint(c) || c == ' ' || c == '\n' || c == '\t'; c = getchar())
;
s = c;
if (c == EOF)
ans = 0;
}
void read1(string &s)
{
s = "";
char c = getchar();
for (; !isprint(c) || c == ' ' || c == '\n' || c == '\t'; c = getchar())
;
for (; isprint(c) && c != ' ' && c != '\n' && c != '\t'; c = getchar())
s += c;
if (c == EOF)
ans = 0;
}
template <typename T>
void read1(T &n)
{
T x = 0;
int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar())
{
if (c == '-')
f = -1;
if (c == EOF)
{
ans = 0;
return;
}
}
for (; isdigit(c); c = getchar())
x = x * 10 + c - 48;
n = x * f;
if (c == EOF)
ans = 0;
if (c != '.')
return;
T l = 0.1;
while ((c = getchar()) <= '9' && c >= '0')
x = x + (c & 15) * l, l *= 0.1;
n = x * f;
if (c == EOF)
ans = 0;
}
void write() {}
void write1(string s)
{
int n = s.size();
for (int i = 0; i < n; i++)
putchar(s[i]);
}
void write1(const char *s)
{
int n = strlen(s);
for (int i = 0; i < n; i++)
putchar(s[i]);
}
void write1(char s) { putchar(s); }
void write1(float s, int x = 6)
{
char y[10001];
sprintf(y, "%%.%df", x);
printf(y, s);
}
void write1(double s, int x = 6)
{
char y[10001];
sprintf(y, "%%.%dlf", x);
printf(y, s);
}
void write1(long double s, int x = 6)
{
char y[10001];
sprintf(y, "%%.%dLf", x);
printf(y, s);
}
template <typename T>
void write1(T n)
{
if (n < 0)
n = -n, putchar('-');
if (n > 9)
write1(n / 10);
putchar('0' + n % 10);
}
template <typename T>
friend ins operator>>(ins x, T &n);
template <typename T>
friend ins operator<<(ins x, T n);
operator bool() { return ans; }
};
template <typename T>
ins operator>>(ins x, T &n)
{
if (!x.ans)
return x;
x.read1(n);
return x;
}
template <typename T>
ins operator<<(ins x, T n)
{
x.write1(n);
return x;
}
ins yin;
ins yout;
#endif
int n;
const int maxn = 4e5;
string s[maxn];
int ss[maxn], b[maxn], bb[maxn];
void pre()
{
for (int i = 1; i <= n; i++)
{
int k = 1e6;
for (int j = 0; j < s[i].size(); j++)
{
if (s[i][j] == 'B')
++b[i];
else
++ss[i];
}
}
}
pair<int, int> ans;
int check(int x)
{
ans = make_pair(0, 0);
int mix = INT_MAX, miy = INT_MAX, maxx = INT_MIN, may = INT_MIN, miz = INT_MAX, maz = INT_MIN;
for (int i = 1; i <= n; i++)
{
mix = min(b[i] + x, mix);
maxx = max(b[i] - x, maxx);
miy = min(ss[i] + x, miy);
may = max(ss[i] - x, may);
miz = min(b[i] - ss[i] + x, miz);
maz = max(b[i] - ss[i] - x, maz);
}
maxx = max(maxx, 0);
may = max(may, 0);
if (mix < maxx)
return 0;
if (miy < may)
return 0;
if (miz < maz)
return 0;
if (maxx - miy > miz || mix - may < maz)
return 0;
ans = make_pair(min(mix, miy + miz), min(min(mix, miy + miz) - maz, miy));
return 1;
}
int main()
{
yin >> n;
for (int i = 1; i <= n; i++)
{
yin >> s[i];
}
pre();
long long l = 0, r = 1e6;
while (l < r)
{
int mid = l + r >> 1;
if (check(mid))
r = mid;
else
l = mid + 1;
}
yout << l << endl;
check(l);
for (int i = 1; i <= ans.first; i++)
yout << 'B';
for (int i = 1; i <= ans.second; i++)
yout << 'N';
return 0;
}
|