AIM Tech Round (Div. 2) C. Graph and String 二分图染色

C. Graph and String

题目连接：

http://codeforces.com/contest/624/problem/C

Description

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 Input

2 1

1 2

Sample Output

Yes

aa

Hint

题意

给你一个图，然后让你给这些图标号，问你有没有一个合法的标号

标号就只能标a,b,c。

然后a会和ab连边，b和abc都连边，c和bc连边

有解就任意输出一个就好

题解：

我们首先把那种边集有n-1的点染成b，然后再随便选一个没有染成b的点当成a，然后和a连边的就是a，没有和a连边的就是c

扫一遍就可以确认所有点的颜色，最后再n^2check一发就好了

代码

#include<bits/stdc++.h>
using namespace std;

int mp[520][520];
int cnt[520];
int flag[520];
int n,m;
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        mp[x][y]=1;
        mp[y][x]=1;
        cnt[x]++;
        cnt[y]++;
    }
    for(int i=1;i<=n;i++)
        if(cnt[i]==n-1)
            flag[i]=2;
    int now=0;
    for(int i=1;i<=n;i++)
        if(flag[i]!=2)
        {
            now=i;
            break;
        }
    if(now==0)
    {
        printf("Yes\n");
        for(int i=1;i<=n;i++)
            cout<<"b";
        cout<<endl;
        return 0;
    }
    flag[now]=1;
    for(int i=1;i<=n;i++)
    {
        if(now==i)continue;
        if(!mp[i][now])flag[i]=3;
        else if(flag[i]==0)flag[i]=1;
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(i==j)continue;
            if(mp[i][j])
            {
                if(abs(flag[j]-flag[i])>1)
                    return puts("No");
            }
            else
            {
                if(abs(flag[j]-flag[i])<=1)
                    return puts("No");
            }
        }
    }
    printf("Yes\n");
    for(int i=1;i<=n;i++)
        if(flag[i]==1)printf("a");
        else if(flag[i]==2)printf("b");
        else printf("c");
    printf("\n");
}