A. Flag

2024-08-28 07:18

文章标签 flag

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A. Flag

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

According to a new ISO standard, a flag of every country should have a chequered field n?×?m, each square should be of one of 10 colours, and the flag should be ?striped?: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets the new ISO standard.

Input

The first line of the input contains numbers n and m (1?≤?n,?m?≤?100), n — the amount of rows, m — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following n lines contain m characters. Each character is a digit between 0 and 9, and stands for the colour of the corresponding square.

Output

Output YES, if the flag meets the new ISO standard, and NO otherwise.

Sample test(s) 
input 
3 3
000
111
222
output 
YES
input 
3 3
000
000
111
output 
NO
input 
3 3
000
111
002
output 
NO
 
 

相邻两行是不同的字符

每行都是一样的字符

/* ***********************************************

************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 1<<30
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
string s;
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n,m;
	while(cin>>n>>m){
		s="";
		string t="";
		int mark=0;
		for(int i=1;i<=n;i++){
			if(i==1){
				cin>>s;
				for(int i=1;i<m;i++)
					if(s[i]!=s[i-1])mark=1;
			}
			else {
				cin>>t;
				if(t==s){
					mark=1;
				}
				if(t!=s){
					for(int i=1;i<m;i++)
						if(t[i]!=t[i-1])mark=1;
					}
				s=t;
			}
		}
		if(mark)cout<<"NO";
		else cout<<"YES";
	}
    return 0;
}