CF1391D-505
题目:
题目描述:
A binary matrix is called good if every even length square sub-matrix has an odd number of ones.
Given a binary matrix $ a $ consisting of $ n $ rows and $ m $ columns, determine the minimum number of cells you need to change to make it good, or report that there is no way to make it good at all.
All the terms above have their usual meanings — refer to the Notes section for their formal definitions.
输入格式:
The first line of input contains two integers $ n $ and $ m $ ( $ 1 \leq n \leq m \leq 10^6 $ and $ n\cdot m \leq 10^6 $ ) — the number of rows and columns in $ a $ , respectively.
The following $ n $ lines each contain $ m $ characters, each of which is one of 0 and 1. If the $ j $ -th character on the $ i $ -th line is 1, then $ a_{i,j} = 1 $ . Similarly, if the $ j $ -th character on the $ i $ -th line is 0, then $ a_{i,j} = 0 $ .
输出格式:
Output the minimum number of cells you need to change to make $ a $ good, or output $ -1 $ if it’s not possible at all.
样例:
样例输入 1:
样例输出 1:
样例输入 2:
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| 7 15
000100001010010
100111010110001
101101111100100
010000111111010
111010010100001
000011001111101
111111011010011
|
样例输出 2:
思路:
实现:
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| /*
*User: ybw051114
*Time: 2020.08.09 22:35:06
*/
#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
const int maxn = 1e6 + 10;
char c[maxn][4];
int f[maxn][8];
int a[10], b[10];
int main()
{
int n, m;
yin >> n >> m;
if (n >= 4 && m >= 4)
{
puts("-1");
return 0;
}
if (n > m)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
yin >> c[i][j];
}
}
swap(n, m);
}
else
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
yin >> c[j][i];
}
}
}
if (n == 1)
{
puts("0");
return 0;
}
for (int i = 0; i < 8; i++)
{
if (i >= 4 && n == 2)
break;
for (int kk = 1; kk <= n; kk++)
if (i & (1 << (kk - 1)))
a[kk] = 1;
else
a[kk] = 0;
int ans = 0;
for (int kk = 1; kk <= n; kk++)
if (a[kk] != c[1][kk] - '0')
ans++;
f[1][i] = ans;
}
for (int k = 2; k <= m; k++)
{
for (int i = 0; i < 8; i++)
{
if (i >= 4 && n == 2)
break;
f[k][i] = INT_MAX;
for (int j = 0; j < 8; j++)
{
if (j >= 4 && n == 2)
break;
for (int kk = 1; kk <= n; kk++)
if (i & (1 << (kk - 1)))
a[kk] = 1;
else
a[kk] = 0;
for (int kk = 1; kk <= n; kk++)
if (j & (1 << (kk - 1)))
b[kk] = 1;
else
b[kk] = 0;
int ans = 0;
for (int kk = 1; kk <= n; kk++)
if (a[kk] != c[k][kk] - '0')
ans++;
if ((a[1] + a[2] + b[1] + b[2]) & 1)
if (n == 2 || ((a[2] + a[3] + b[2] + b[3]) & 1))
f[k][i] = min(f[k][i], f[k - 1][j] + ans);
}
}
}
if (n == 2)
{
int ans = INT_MAX;
for (int i = 0; i < 4; i++)
ans = min(ans, f[m][i]);
yout << ans << endl;
return 0;
}
int ans = INT_MAX;
for (int i = 0; i < 8; i++)
ans = min(ans, f[m][i]);
yout << ans << endl;
return 0;
}
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