Codeforces AIM Tech Round (Div. 1)623A Graph and String

A. Graph and String

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

One day student Vasya was sitting on a lecture and mentioned a string s1s2… sn, consisting of letters “a”, “b” and “c” that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:

G has exactly n vertices, numbered from 1 to n.

For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs “a”-“b” and “b”-“c” are neighbouring, while letters “a”-“c” are not.

Vasya painted the resulting graph near the string and then erased the string. Next day Vasya’s friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya’s adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.

Input

The first line of the input contains two integers n and m — the number of vertices and edges in the graph found by Petya, respectively.

Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.

Output

In the first line print “Yes” (without the quotes), if the string s Petya is interested in really exists and “No” (without the quotes) otherwise.

If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters “a”, “b” and “c” only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.

Sample test(s)

input

2 1

1 2

output

Yes

aa

input

4 3

1 2

1 3

1 4

output

No

Note

In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings “aa”, “ab”, “ba”, “bb”, “bc”, “cb”, “cc” meets the graph’s conditions.

In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.

题意：

很好理解的题目，大意就是说，在一个只含有a,b,c三个字母的无向图中，根据给出的点数n与边数m以及点与点之间的相互连通关系，问是否存在这样一串字符串，无则输出No，有则输出Yes并给出答案。

思路：

题目是在比赛结束后看到别人的代码才恍然大悟的，比赛过程中一直dfs结果各种无从de起的bug让我心碎(:з」∠)。。根据条件以及图形构造，我们可以知道如果一个点与其他点都有连通，那么这个点可以设为b，其他点为a或c，从第一个点开始遍历，如果一个点不是b，那么设为a，其他与之没有连通的点设为c。在一个构造出来的图中，如果两个点之间有连线，那么必然是a-b或者是b-c，如果是a-c，则矛盾，输出no；还有另外一个判断，如果两个点之间没有连线而两个点必然为a-c，否则矛盾。

代码如下：

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,m;
while(cin>>n>>m)
{
int str[550][550],cnt[550],ans[550];
memset(str,0,sizeof(str));
memset(cnt,0,sizeof(cnt));
memset(ans,0,sizeof(ans));
for(int i = 0; i < m; i++)
{
int u,v;
cin>>u>>v;
str[u][v] = str[v][u] = 1;
cnt[u]++;
cnt[v]++;
}
for(int i = 1; i <= n; i++)
if(cnt[i] == n - 1)
ans[i] = 1;
for(int i = 1; i <= n; i++)
if(ans[i] == 0)
for(int j = i + 1; j <= n; j++)
if(str[i][j] == 0)
ans[j] = 2;
bool flag = true;
for(int i = 1; i <= n; i++)
{
for(int j = i + 1; j <= n; j++)
{
if((abs(ans[i] - ans[j]) == 2 && str[i][j]) || (abs(ans[i] - ans[j]) < 2 && !str[i][j]))
{
cout<<"No"<<endl;
flag = false;
break;
}
}
if(!flag)
break;
}
if(!flag)
continue;
cout<<"Yes"<<endl;
for(int i = 1; i <= n; i++)
printf("%c",ans[i] + 'a');
cout<<endl;
}
return 0;
}