Knights of the Round Table
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 8785 | Accepted: 2813 |
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.
题意不说了，经典问题。
解题思路:主要是奇圈的判断，由于二分图与奇圈恰好相对，所以对每一个圈判断二分图，最后统计即可。
代码：
/* ***********************************************
Author :xianxingwuguan
Created Time :2014/1/18 18:00:02
File Name :F.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int maxn=3000;
int head[maxn],dfn[maxn],mark[maxn],odd[maxn],Stack[maxn],tol,indexx,top,col[maxn],low[maxn];
int n;
struct node
{
int next,to;
}edge[1000000];
void add(int u,int v)
{
edge[tol].to=v;
edge[tol].next=head[u];
head[u]=tol++;
}
bool judge(int s)
{
memset(col,-1,sizeof(col));
int cnt=0;
for(int i=1;i<=n;i++)cnt+=mark[i];
if(cnt<3)return 1;
queue<int> q;
q.push(s);
col[s]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(!mark[v])continue;
if(col[v]==col[u])
return 0;
if(col[v]==-1)
{
col[v]=1-col[u];
q.push(v);
}
}
}
return 1;
}
void fun(int u)
{
if(judge(u)==0)
{
for(int i=1;i<=n;i++)
if(mark[i])
odd[i]=1;
}
}
void tarjin(int u,int pre)
{
low[u]=dfn[u]=++indexx;
Stack[top++]=u;
int v;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(v==pre)continue;
if(!dfn[v])
{
tarjin(v,u);
if(low[u]>low[v])
low[u]=low[v];
if(low[v]>=dfn[u])
{
memset(mark,0,sizeof(mark));
int vn;
do
{
vn=Stack[--top];
mark[vn]=1;
}while(vn!=v);
mark[u]=1;
fun(u);
}
}
else if(low[u]>dfn[v])
low[u]=dfn[v];
}
}
int mp[2000][2000];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int i,j,k,m;
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0)break;
memset(head,-1,sizeof(head));tol=0;
memset(dfn,0,sizeof(dfn));
memset(odd,0,sizeof(odd));
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
mp[i][j]=1;
while(m--)
{
scanf("%d%d",&i,&j);
mp[i][j]=mp[j][i]=0;
}
for(i=1;i<=n;i++)
for(j=i+1;j<=n;j++)
if(mp[i][j])
add(i,j),add(j,i);
indexx=top=0;
for(i=1;i<=n;i++)
if(!dfn[i])
tarjin(i,i);
int ans=0;
for(i=1;i<=n;i++)
if(!odd[i])
ans++;
printf("%d\n",ans);
}
return 0;
}