【POJ 2942】Knights of the Round Table（双联通分量+染色判奇环）
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 11661 | Accepted: 3824 |
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
Hint
Huge input file, 'scanf' recommended to avoid TLE.
Source
题目大意：有n个骑士，m个仇视关系，每一个关系a b表示a与b老死不相往来，。
如今亚瑟王想要定期举办圆桌会议。圆桌会议要求到的骑士围成一圈。也就是每一个骑士一定会有左右两个相邻的骑士。要求相邻的骑士间不可有直接的仇视关系。
举行圆桌会议的骑士数目必须大于1。而且必须为奇数。
问须要剔除多少骑士才干正常举办会议。
有一个意思我没读出来，就是并不须要留下的骑士能參与同一场会议，会议能够举行多场，仅仅要留下的骑士能參与当中一场就可以。
这样建立补图，也就是骑士友好关系的图后。处于不同的双连通子图中的点一定无法出席同一场圆桌会议。由于这些点间要么是仇恨关系，要么仅仅有一条友好关系，无法成环。
这样范围就缩小到同一双连通分量中。
这里用到了一个结论，对于一个双连通分量，假设存在奇环，那么这个双连通分量中的每个点都一定会存在于至少一个奇环中。
由于不论什么一个点都有两条以上到这个奇环的路径，两条路径在连接奇环上的两个点，能够把奇环变为偶链和奇链 这样依据该点到这两个点间点个数的奇偶性进行选择 就保证一定能够构成奇环。
相同，假设不存在奇环，则全部双连通分量中的点都不存在于不论什么一个奇环中。
代码例如以下：
#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)
using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;
bool can[2333];
bool in[2333];
bool mp[2333][2333];
bool vis[2333];
int col[2333];
int dfn[2333],low[2333];
stack <int> s;
int n,tim;
bool cal(int u)
{
// printf("%d
",u);
queue <int> q;
q.push(u);
memset(col,-1,sizeof(col));
col[u] = 1;
while(!q.empty())
{
u = q.front();
q.pop();
for(int i = 1; i <= n; ++i)
{
if(u != i && in[i] && !mp[u][i])
{
// printf("%d->%d %d %d
",u,i,col[u],col[i]);
if(col[i] == -1)
{
col[i] = col[u]^1;
q.push(i);
}else if(col[i] == col[u]) return true;
}
}
}
return false;
}
void tarjan(int u,int pre)
{
s.push(u);
dfn[u] = low[u] = tim++;
vis[u] = 1;
for(int i = 1; i <= n; ++i)
{
if(i == u || i == pre || mp[u][i]) continue;
if(!vis[i])
{
tarjan(i,u);
low[u] = min(low[u],low[i]);
if(low[i] >= dfn[u])
{
memset(in,0,sizeof(in));
while(s.top() != i)
{
in[s.top()] = 1;
s.pop();
}
in[i] = 1;
s.pop();
in[u] = 1;
if(cal(u))
{
for(int i = 1; i <= n; ++i)
if(in[i]) can[i] = 1;
}
}
}else low[u] = min(low[u],dfn[i]);
}
}
int main()
{
//fread();
//fwrite();
int m,u,v;
while(~scanf("%d%d",&n,&m) && (n+m))
{
memset(mp,0,sizeof(mp));
while(m--)
{
scanf("%d%d",&u,&v);
mp[u][v] = mp[v][u] = 1;
}
memset(vis,0,sizeof(vis));
memset(can,0,sizeof(can));
tim = 0;
for(int i = 1; i <= n; ++i)
if(!vis[i]) tarjan(i,i);
int ans = 0;
for(int i = 1; i <= n; ++i)
ans += can[i];
printf("%d
",n-ans);
}
return 0;
}