【POJ】2942 Knights of the Round Table 点双连通经典题

Knights of the Round Table

Time Limit: 7000MS
Memory Limit: 65536KTotal Submissions: 9411
Accepted: 3045

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:

• 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.)

Merlin will let the knights sit down only if these two rules are satisfied, otherwise he cancels the meeting. (If only one knight shows up, then the meeting is canceled as well, as one person cannot sit around a table.) Merlin realized that this means that there can be knights who cannot be part of any seating arrangements that respect these rules, and these knights will never be able to sit at the Round Table (one such case is if a knight hates every other knight, but there are many other possible reasons). If a knight cannot sit at the Round Table, then he cannot be a member of the Knights of the Round Table and must be expelled from the order. These knights have to be transferred to a less-prestigious order, such as the Knights of the Square Table, the Knights of the Octagonal Table, or the Knights of the Banana-Shaped Table. To help Merlin, you have to write a program that will determine the number of knights that must be expelled.

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 .

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 51 41 52 53 44 50 0

Sample Output

2

Hint

Huge input file, 'scanf' recommended to avoid TLE.

Source

Central Europe 2005

传送门：【POJ】2942 Knights of the Round Table

题目大意：
有n个骑士经常举行圆桌会议，每次圆桌会议至少应该有3个骑士参加，且相互憎恨的骑士不能坐在圆桌旁的相邻位置。如果发生意见分歧，则需要举手表决，因此参加会议的骑士数目必须是奇数，以防赞同和反对票一样多。现在你知道哪些骑士相互憎恨后，你的任务是统计有多少个骑士不可能参加任何一个会议。

题目分析：假设不能相邻的骑士之间有一条边，以此构成图G，那么能够相邻的骑士之间建边构成G的补图G'，则题目转化成求不在任何一个简单奇圈上的节点数。如果图G不连通，应该对每个连通分量求解。
简单圈上的点必定属于同一个双连通分量，因此需要找到所有的双连通分量。因为二分图是没有奇圈的（因为总是能够做到黑白染色），所以我们只需要找不是二分图的双连通分量。怎么找？黑白染色判定，如果一个连通分量可以被黑白染色（即相邻两点颜色不同），则说明该双连通分量是一个二分图，否则该双连通分量不是二分图（必定包含奇圈）。
我们只需要在每得到一个连通块时就进行染色判定即可。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define up( i ) ( ( i ) << 1 )#define down( i ) ( ( i ) << 1 | 1 )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REPV( i , a , b ) for ( int i = a ; i >= b ; -- i )#define clear( a , x ) memset ( a , x , sizeof a )const int MAXN = 1005 ;const int MAXE = 1000005 ;const int INF = 0x7f7f7f7f ;struct Edge {int v , n ;Edge ( int var = 0 , int next = 0 ) :v ( var ) , n ( next ) {}} ;struct BCC {Edge edge[MAXE] ;int adj[MAXN] , cntE ;int bcc[MAXN] , bcc_cnt ;int low[MAXN] , dfn[MAXN] , dfs_clock ;int S[MAXN] , top ;int n , m ;int color[MAXN] ;bool G[MAXN][MAXN] ;bool odd[MAXN] ;void init () {cntE = top = bcc_cnt = dfs_clock = 0 ;clear ( G , 0 ) ;clear ( bcc , 0 ) ;clear ( odd , 0 ) ;clear ( dfn , 0 ) ;clear ( adj , -1 ) ;}void addedge ( int u , int v ) {edge[cntE] = Edge ( v , adj[u] ) ;adj[u] = cntE ++ ;edge[cntE] = Edge ( u , adj[v] ) ;adj[v] = cntE ++ ;}int draw ( int u , int co ) {for ( int i = adj[u] ; ~i ; i = edge[i].n ) {int v = edge[i].v ;if ( bcc[v] != co )continue ;if ( !color[v] ) {color[v] = 3 - color[u] ;if ( !draw ( v , co ) )return 0 ;}else if ( color[v] != 3 - color[u] )return 0 ;}return 1 ;}void tarjan ( int u , int fa ) {low[u] = dfn[u] = ++ dfs_clock ;S[top ++] = u ;int flag = 1 ;for ( int i = adj[u] ; ~i ; i = edge[i].n ) {int v = edge[i].v ;if ( v == fa && flag ) {flag = 0 ;continue ;}if ( !dfn[v] ) {tarjan ( v , u ) ;low[u] = min ( low[u] , low[v] ) ;if ( low[v] >= dfn[u] ) {++ bcc_cnt ;while ( 1 ) {int x = S[-- top] ;bcc[x] = bcc_cnt ;if ( x == v )break ;}bcc[u] = bcc_cnt ;clear ( color , 0 ) ;color[u] = 1 ;if ( !draw ( u , bcc_cnt ) )REPF ( j , 1 , n )if ( bcc[j] == bcc_cnt )odd[j] = 1 ;}}elselow[u] = min ( low[u] , dfn[v] ) ;}}void input () {int u , v ;REP ( i , m ) {scanf ( "%d%d" , &u , &v ) ;G[u][v] = G[v][u] = 1 ;}REPF ( i , 1 , n )REPF ( j , i + 1 , n )if ( !G[i][j] )addedge ( i , j ) ;}void solve () {int cnt = 0 ;REPF ( i , 1 , n )if ( !dfn[i] )tarjan ( i , -1 ) ;REPF ( i , 1 , n )if ( !odd[i] )++ cnt ;printf ( "%d\n" , cnt ) ;}} ;BCC z ;void work () {int n , m ;while ( ~scanf ( "%d%d" , &z.n , &z.m ) && ( z.n || z.m ) ) {z.init () ;z.input () ;z.solve () ;}}int main () {work () ;return 0 ;}