AW365 圆桌骑士

题目地址

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简述：

• 缩点+二分图染色（判奇环）

易错点：

• 为了防止错误加边（自环，即i==j），可将加边时的第二层遍历从i+1开始，由对称性可知加边无错误.
• vector直接用[]比at()更美观.
• 对于每个dcc都要重置一次color，这是因为同一个点可能属于多个dcc.

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#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
const int MAXN=2e3,MAXM=2e6;
struct Edge{
	int from,to,w,nxt;
}e[MAXM];
int head[MAXN],edgeCnt=1;
void addEdge(int u,int v,int w){
	e[++edgeCnt].from=u;
	e[edgeCnt].to=v;
	e[edgeCnt].w=w;
	e[edgeCnt].nxt=head[u];
	head[u]=edgeCnt;
}
int dfn[MAXN],dfnCnt=0,low[MAXN];
vector<int> dcc[MAXN];
int dccCnt=0,c[MAXN];//in which dcc
int stck[MAXN],top=0;
void tarjan(int x,int root){
	dfn[x]=low[x]=++dfnCnt;
	stck[++top]=x;
	if(x==root&&(!head[x])){
		dcc[++dccCnt].push_back(x);
		return;
	}
	for(int i=head[x];i;i=e[i].nxt){
		int nowV=e[i].to;
		if(!dfn[nowV]){
			tarjan(nowV,root);
			low[x]=min(low[x],low[nowV]);
			if(low[nowV]>=dfn[x]){
				int y;dccCnt++;
				do{
					y=stck[top--];
					dcc[dccCnt].push_back(y);
				}while(y!=nowV);
				dcc[dccCnt].push_back(x);
			}
		}else low[x]=min(low[x],dfn[nowV]);
	}
}
bool canAttend[MAXN],flag=0;
int color[MAXN];
void dfs(int x,int restColor){
	color[x]=restColor;
	for(int i=head[x];i;i=e[i].nxt){
		int nowV=e[i].to;
		if(c[nowV]!=c[x])continue;
		if(!color[nowV])dfs(nowV,3-restColor);
		else if(color[nowV]==color[x]){
			flag=1;
			return;
		}
	}
}
bool hate[MAXN][MAXN];
int n;
void init(){
	memset(head,0,sizeof(head));
	edgeCnt=1;
	memset(hate,0,sizeof(hate));
	memset(dfn,0,sizeof(dfn));
	dfnCnt=0;
	memset(low,0,sizeof(low));
	for(int i=1;i<=n;i++){
		dcc[i].clear();
	}
	dccCnt=0;
	memset(c,0,sizeof(c));
	memset(canAttend,0,sizeof(canAttend));
	top=0;
}
int main(){
	while(1){
		init();
		int m;
		scanf("%d%d",&n,&m);
		if(!n||!m)break;
		for(int i=1;i<=m;i++){
			int a,b;
			scanf("%d%d",&a,&b);
			hate[a][b]=hate[b][a]=1;
		}
		for(int i=1;i<n;i++){//
			for(int j=i+1;j<=n;j++){
				if(!hate[i][j]){
					addEdge(i,j,1);
					addEdge(j,i,1);
				}
			}
		}
		for(int i=1;i<=n;i++){
			if(!dfn[i])tarjan(i,i);
		}
		for(int i=1;i<=dccCnt;i++){
			for(int j=0;j<dcc[i].size();j++){
				c[dcc[i][j]]=i;//
				color[dcc[i][j]]=0;//
			}
			flag=0;
			dfs(dcc[i][0],1);
			if(flag){//can attend
				for(int j=0;j<dcc[i].size();j++){
					canAttend[dcc[i][j]]=1;
				}
			}
		}
		int ans=0;
		for(int i=1;i<=n;i++){
			if(!canAttend[i])ans++;
		}
		printf("%d
",ans);
	}
	return 0;
}