新博客地址: vonsdite.cn
题目网址: Educational Codeforces Round 26 B. Flag of Berland
题意分析:
题意:
问这个旗子 满不满足一下条件
有三条线, 且三条线颜色不一样, 宽度, 高度, 都相等, 且平行每一条线只有一种颜色
思路:
暴力判断即可从行的角度来判断是否是三条线, 是否每条线等宽等高,是否颜色不一样从列的角度同行
代码:
#include <bits/stdc++.h>
using namespace std;
const int SIZE = 105;
char flag[SIZE][SIZE];
int n, m;
bool judge_hang()
{
int color[200];
memset(color, 0, sizeof(color));
int t1 = n / 3, t2 = n * 2 / 3, t3 = n;
char tmpColor = flag[0][0];
color[(int)tmpColor] = 1;
for (int i = 0; i < t1; ++i)
{
for (int j = 0; j < m; ++j)
{
if(flag[i][j] != tmpColor)
{
return false;
}
}
}
tmpColor = flag[t1][0];
if(color[(int)tmpColor]) return false;
else
{
color[(int)tmpColor] = 1;
for (int i = t1; i < t2; ++i)
{
for (int j = 0; j < m; ++j)
{
if(flag[i][j] != tmpColor)
{
return false;
}
}
}
}
tmpColor = flag[t2][0];
if(color[(int)tmpColor]) return false;
else
{
for (int i = t2; i < t3; ++i)
{
for (int j = 0; j < m; ++j)
{
if(flag[i][j] != tmpColor)
{
return false;
}
}
}
}
return true;
}
bool judge_lie()
{
int color[200];
int t1 = m / 3, t2 = m * 2 / 3, t3 = m;
memset(color, 0, sizeof(color));
char tmpColor = flag[0][0];
color[(int)tmpColor] = 1;
for (int i = 0; i < t1; ++i)
{
for (int j = 0; j < n; ++j)
{
if(flag[j][i] != tmpColor)
{
return false;
}
}
}
tmpColor = flag[0][t1];
if(color[(int)tmpColor]) return false;
else
{
color[(int)tmpColor] = 1;
for (int i = t1; i < t2; ++i)
{
for (int j = 0; j < n; ++j)
{
if(flag[j][i] != tmpColor)
{
return false;
}
}
}
}
tmpColor = flag[0][t2];
if(color[(int)tmpColor]) return false;
else
{
color[(int)tmpColor] = 1;
for (int i = t2; i < t3; ++i)
{
for (int j = 0; j < n; ++j)
{
if(flag[j][i] != tmpColor)
{
return false;
}
}
}
}
return true;
}
int main(int argc, char const *argv[])
{
scanf("%d %d", &n, &m);
getchar();
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < m; ++j)
{
scanf("%c", &flag[i][j]);
}
getchar();
}
if(n % 3 && m % 3)
{
printf("NO\n");
return 0;
}
else
{
if(n % 3 == 0)
{
if(judge_hang())
{
printf("YES\n");
return 0;
}
}
if(m % 3 == 0)
{
if(judge_lie())
{
printf("YES\n");
return 0;
}
}
printf("NO\n");
}
return 0;
}
