[POJ2942][LA3523]Knights of the Round Table

试题描述

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.

(n) 个骑士经常参加圆桌会议，一次会议需要选多于一个骑士坐在一张圆桌旁，并且需要满足：

互相憎恨的骑士不能坐在相邻位置；
圆桌旁有奇数个骑士。

问有多少个骑士不能参加任何会议。

输入

The input contains several blocks of test cases. Each case begins with a line containing two integers (1 le n le 1000) and (1 le m le 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 (k_1) and (k_2) , which means that knight number (k_1) and knight number (k_2) hate each other (the numbers (k_1) and (k_2) are between (1) and (n) ).

The input is terminated by a block with n = m = 0 .

输出

For each test case you have to output a single integer on a separate line: the number of knights that have to be expelled.

输入示例

输出示例

数据规模及约定

见“输入”

题解

注意这题是点-双连通分量。（判断是边双还是点双，画几个“8 字形”就好了）

我们将所有的点双依次处理（割点一定要被处理多次，它属于多个点双），看每个点双是否能被二分图染色，不能被二分染色的点双中所有点是可以参加圆桌会议的。

我们可以证明如果一个点双中含有一个简单奇环，那么该双连通分量中所有点都可以在某个简单奇环中。读者不妨自己画一个有交（指边同时在两个环中）奇环和偶环，发现偶环上所有点都可以将它所在偶环的那一半和奇环的一半拼出一个大奇环。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <vector>
using namespace std;
#define rep(i, s, t) for(int i = (s); i <= (t); i++)
#define dwn(i, s, t) for(int i = (s); i >= (t); i--)

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
	if(Head == Tail) {
		int l = fread(buffer, 1, BufferSize, stdin);
		Tail = (Head = buffer) + l;
	}
	return *Head++;
}
int read() {
	int x = 0, f = 1; char c = Getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }
	return x * f;
}

#define maxn 1010
#define maxm 2000010

struct Edge {
	int a, b;
	Edge() {}
	Edge(int _, int __): a(_), b(__) {}
} es[maxm];
int n, m, head[maxn], nxt[maxm];
bool G[maxn][maxn];

void AddEdge(int a, int b) {
	es[++m] = Edge(a, b); nxt[m] = head[a]; head[a] = m;
	swap(a, b);
	es[++m] = Edge(a, b); nxt[m] = head[a]; head[a] = m;
	return ;
}

int clo, dfn[maxn], low[maxn], bcno[maxn], cntb, top;
Edge S[maxm];
vector <int> bcc[maxn];
bool iscut[maxn];
void dfs(int u, int fa) {
	dfn[u] = low[u] = ++clo;
	int ch = 0;
	for(int i = head[u]; i; i = nxt[i]) if(i != fa) {
		Edge& e = es[i];
		if(dfn[e.b]) low[u] = min(low[u], dfn[e.b]);
		else {
			S[++top] = e;
			dfs(e.b, i); ch++;
			low[u] = min(low[u], low[e.b]);
			if(low[e.b] >= dfn[u]) {
				iscut[u] = 1;
				cntb++;
				while(1) {
					Edge s = S[top--];
					if(bcno[s.a] != cntb) bcno[s.a] = cntb, bcc[cntb].push_back(s.a);
					if(bcno[s.b] != cntb) bcno[s.b] = cntb, bcc[cntb].push_back(s.b);
					if(s.a == e.a && s.b == e.b) break;
				}
			}
		}
	}
	if(!fa && ch == 1) iscut[u] = 0;
	return ;
}

bool cant[maxn];
int col[maxn];
bool solve(int u, int b, int c) {
	if(col[u]) return col[u] == c;
	col[u] = c;
	for(int i = head[u]; i; i = nxt[i]) {
		Edge& e = es[i];
		if(bcno[e.b] == bcno[u] && !solve(e.b, b, 3 - c)) return 0;
	}
	return 1;
}

int main() {
	while(1) {
		n = read(); int M = read();
		if(!n) break;
		
		memset(G, 0, sizeof(G));
		rep(i, 1, M) {
			int a = read(), b = read();
			G[a][b] = G[b][a] = 1;
		}
		
		m = 0; memset(head, 0, sizeof(head));
		rep(i, 1, n)
			rep(j, i + 1, n) if(!G[i][j]) AddEdge(i, j);
		
		memset(dfn, 0, sizeof(dfn)); clo = 0;
		memset(low, 0, sizeof(low));
		memset(bcno, 0, sizeof(bcno));
		rep(i, 1, cntb) bcc[i].clear();
		cntb = 0;
		rep(i, 1, n) if(!dfn[i]) dfs(i, 0);
		/*printf("cntb: %d
", cntb);
		rep(i, 1, cntb) {
			printf("%d:", i);
			for(vector <int> :: iterator j = bcc[i].begin(); j != bcc[i].end(); j++)
				printf(" %d", *j);
			putchar('
');
		}*/
		
		memset(cant, 0, sizeof(cant));
		rep(i, 1, cntb) {
			for(vector <int> :: iterator j = bcc[i].begin(); j != bcc[i].end(); j++)
				bcno[*j] = i;
			memset(col, 0, sizeof(col));
			if(!solve(*bcc[i].begin(), i, 1))
				for(vector <int> :: iterator j = bcc[i].begin(); j != bcc[i].end(); j++)
					cant[*j] = 1;
		}
		
		int cnt = n;
		rep(i, 1, n) cnt -= cant[i];
		printf("%d
", cnt);
	}
	
	return 0;
}