Description
Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the other knights are fun things to do. Therefore, it is not very surprising that in recent years the kingdom of King Arthur has experienced an unprecedented increase in the number of knights. There are so many knights now, that it is very rare that every Knight of the Round Table can come at the same time to Camelot and sit around the round table; usually only a small group of the knights isthere, while the rest are busy doing heroic deeds around the country.
Knights can easily get over-excited during discussions-especially after a couple of drinks. After some unfortunate accidents, King Arthur asked the famous wizard Merlin to make sure that in the future no fights break out between the knights. After studying the problem carefully, Merlin realized that the fights can only be prevented if the knights are seated according to the following two rules:
Knights can easily get over-excited during discussions-especially after a couple of drinks. After some unfortunate accidents, King Arthur asked the famous wizard Merlin to make sure that in the future no fights break out between the knights. After studying the problem carefully, Merlin realized that the fights can only be prevented if the knights are seated according to the following two rules:
- The knights should be seated such that two knights who hate each other should not be neighbors at the table. (Merlin has a list that says who hates whom.) The knights are sitting around a roundtable, thus every knight has exactly two neighbors.
- An odd number of knights should sit around the table. This ensures that if the knights cannot agree on something, then they can settle the issue by voting. (If the number of knights is even, then itcan happen that ``yes" and ``no" have the same number of votes, and the argument goes on.)
Input
The input contains several blocks of test cases. Each case begins with a line containing two integers 1 ≤ n ≤ 1000 and 1 ≤ m ≤ 1000000 . The number n is the number of knights. The next m lines describe which knight hates which knight. Each of these m lines contains two integers k1 and k2 , which means that knight number k1 and knight number k2 hate each other (the numbers k1 and k2 are between 1 and n ).
The input is terminated by a block with n = m = 0 .
The input is terminated by a block with n = m = 0 .
Output
For each test case you have to output a single integer on a separate line: the number of knights that have to be expelled.
Sample Input
5 5
1 4
1 5
2 5
3 4
4 5
0 0
Sample Output
2
题目大意：有n个骑士，他们之间可能有憎恨关系，例如：骑士a 憎恨 骑士b ，那么骑士b 也一定 憎恨 骑士a。现在要给这n个骑士开会，会议需要满足以下要求：
1、参加会议的骑士数量必须是奇数。
2、会议选用圆形桌子，即骑士们开会时围成一个圈。
3、要求开会时任意一个骑士与相邻的两个骑士之间没有憎恨关系。
当然，会议可以开很多场，而且一个骑士也可以参加很多场会议（如果能参加的话），问：一场会议也不能参加的骑士数量是多少？
解题思路：
1、先说一下建图方法：将两个没有憎恨关系的骑士之间连接一条边（无向边），代表开会时这两个骑士可以相邻。
2、判断一个骑士能否参加会议，就是判断在这个无向图中有没有存在一个简单的奇圈包含这个骑士（简单奇圈是指由奇数个顶点组成的圈，并且这些顶点互不相同）。
3、在同一个简单奇圈上的点必然在同一个双连通分量中，所以要找出图中所有的双连通分量，然后判断这个双连通分量中有没有奇圈。二分图中是没有奇圈的，如果一个双联通分量不是二分图，那么一定存在奇圈（此处不再证明），所以只需判断一个双连通分量是不是二分图即可，如果是二分图，那么这个连通分量中每个点都可以参加会议。
请看代码：
#include<iostream>
#include<string>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#define mem(a , b) memset(a , b , sizeof(a))
using namespace std ;
inline void RD(int &a)
{
a = 0 ;
char t ;
do
{
t = getchar() ;
}
while (t < '0' || t > '9') ;
a = t - '0' ;
while ((t = getchar()) >= '0' && t <= '9')
{
a = a * 10 + t - '0' ;
}
}
inline void OT(int a)
{
if(a >= 10)
{
OT(a / 10) ;
}
putchar(a % 10 + '0') ;
}
const int MAXN = 1005 ;
typedef struct edge
{
int u ;
int v ;
} E ;
vector<int>vert[MAXN] ;
vector<int>bscnt[MAXN] ; // 记录每个双连通分量里的顶点
int blcnt[MAXN] ; // 记录每个顶点属于哪个连通分量
bool ha[MAXN][MAXN] ; // 标记两个骑士是否相互憎恨
bool vis[MAXN] ;
int n , m ;
int dfn[MAXN] ;
int low[MAXN] ;
int tmpdfn ;
int top ;
E stap[MAXN * MAXN] ;
bool odd[MAXN] ;
short color[MAXN] ;
int scnt ;
void clr()
{
mem(ha , 0) ;
mem(dfn , 0) ;
mem(low , 0) ;
mem(vis , 0) ;
mem(blcnt , 0) ;
mem(color , 0) ;
mem(odd , 0) ;
int i ;
for(i = 0 ; i <= n ; i ++)
{
vert[i].clear() ;
bscnt[i].clear() ;
}
top = -1 ;
tmpdfn = 0 ;
scnt = 0 ;
}
void tarjan(int u , int fa)
{
int son = 0 ;
vis[u] = true ;
dfn[u] = low[u] = ++ tmpdfn ;
int i ;
for(i = 0 ; i < vert[u].size() ; i ++)
{
int v = vert[u][i] ;
E e ;
e.u = u ;
e.v = v ;
if(!vis[v])
{
stap[++ top] = e ;
son ++ ;
tarjan(v , u) ;
low[u] = min(low[u] , low[v]) ;
if(low[v] >= dfn[u])
{
scnt ++ ;
int tu , tv ;
while (1)
{
E tmp = stap[top --] ;
tu = tmp.u ;
tv = tmp.v ;
if(blcnt[tu] != scnt) // 注意此处的条件，由于关节点属于
//不同的连通分量，所以条件不能写成(!blcnt[tu])
{
blcnt[tu] = scnt ;
bscnt[scnt].push_back(tu) ;
}
if(blcnt[tv] != scnt)
{
blcnt[tv] = scnt ;
bscnt[scnt].push_back(tv) ;
}
if(tu == u && tv == v)
break ;
}
}
}
else if(v != fa && dfn[v] < dfn[u]) // 注意此处的判断条件，不要漏掉 dfn[v] < dfn[u]
{
stap[++ top] = e ;
low[u] = min(low[u] , dfn[v]) ;
}
}
}
void init()
{
clr() ;
int i , j ;
for(i = 0 ; i < m ; i ++)
{
int a , b ;
RD(a) ;
RD(b) ;
ha[a][b] = ha[b][a] = true ;
}
for(i = 1 ; i <= n ; i ++) // 建图
{
for(j = 1 ; j <= n ; j ++)
{
if(!ha[i][j] && i != j)
vert[i].push_back(j) ;
}
}
}
bool isbg(int u , int c) //判断是否为二分图
{
int i ;
for(i = 0 ; i < vert[u].size() ; i ++)
{
int v = vert[u][i] ;
if(blcnt[v] != c)
continue ;
if(color[v] == color[u])
return false ;
if(!color[v])
{
color[v] = 3 - color[u] ;
if(!isbg(v , c))
return false ;
}
}
return true ;
}
void solve()
{
mem(vis , 0) ;
int i ;
for(i = 1 ; i <= n ; i ++)
{
if(!vis[i])
{
tarjan(i , -1) ;
}
}
for(i = 1 ; i <= scnt ; i ++)
{
int j ;
int tmp ;
mem(color , 0) ;
for(j = 0 ; j < bscnt[i].size() ; j ++)
{
tmp = bscnt[i][j] ;
blcnt[tmp] = i ;
}
color[tmp] = 1 ;
if(!isbg(tmp , i))
{
for(j = 0 ; j < bscnt[i].size() ; j ++)
{
tmp = bscnt[i][j] ;
odd[tmp] = true ;
}
}
}
int ans = n ;
for(i = 1 ; i <= n ; i ++)
{
if(odd[i])
ans -- ;
}
OT(ans) ;
puts("") ;
}
int main()
{
while (scanf("%d%d" , &n , &m) != EOF)
{
if(n == 0 && m == 0)
break ;
init() ;
solve() ;
}
return 0 ;
}