2020 ICPC上海站 B

2020 ICPC上海站 B

题目链接：https://ac.nowcoder.com/acm/contest/9925/B

题目大意

给两个\(n*m\)的扫雷图，要求对第二个图修改不超过\((n*m)/2\)次，使得图中无雷区数字之和等于第一个图的无雷区数字之和。

思路

水题,一个扫雷图数字之和和他的翻转图是相同的，两个图至少有一个相似度超过50%，输出即可。

Code

#include<string>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define LL long long
#define MOD 100000007
#define PI 3.1415926535898
#define INF 0x3f3f3f3f
#define MAXN 200050
#define register long long
const double EPS = 1e-8;
LL read()
{
	LL w = 1, x = 0;
	char ch = 0;
	while (ch < '0' || ch>'9')
	{
		if (ch == '-')
			w = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9')
	{
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return w * x;
}
LL n, m;
LL mp[1005][1005];
string s;
int main()
{
	n = read();
	m = read();
	for (register int i = 1; i <= n; i++)
	{
		cin >> s;
		for (register int j = 0; j < m; j++)
		{
			if (s[j] == '.')
			{
				mp[i][j + 1] = 1;
			}
		}
	}
	int x = 0;
	for (register int i = 1; i <= n; i++)
	{
		cin >> s;
		for (register int j = 0; j < m; j++)
		{
			if (s[j] == '.')
			{
				if (mp[i][j + 1] == 1) x++;
			}
			else
			{
				if (mp[i][j + 1] == 0) x++;
			}
		}
	}
	if (x > n * m / 2)
	{
		for (register int i = 1; i <= n; i++)
		{
			for (register int j = 1; j <= m; j++)
			{
				if (mp[i][j])
				{
					cout << '.';
				}
				else
				{
					cout << 'X';
				}
			}
			cout << endl;
		}
	}
	else
	{
		for (register int i = 1; i <= n; i++)
		{
			for (register int j = 1; j <= m; j++)
			{
				if (mp[i][j])
				{
					cout << 'X';
				}
				else
				{
					cout << '.';
				}
			}
			cout << endl;
		}
	}
	return 0;
}