【tarjan求双连通分量+染色判二分图】POJ 2942

#define N 1005
vector<int> v
;
vector<int> lin
;
int dfn
,low
;
bool instack
,iscut
;
int num_rt;
int cnt;
int step;
stack<int> sta;
bool g

;
void init(int n){
for(int i=0;i<=n;i++){
instack[i] = 0;
iscut[i] = 0;
dfn[i] = 0;
low[i] = 0;
}
step = 0;
cnt = 0;
while(!sta.empty())sta.pop();
}
void tarjan(int u,int fa){
low[u] = dfn[u] = ++step;
sta.push(u);
instack[u] = 1;
int i;
for(i=0;i<v[u].size();i++){
int to = v[u][i];
if(to==fa)continue;
if(!dfn[to]){
tarjan(to,u);
low[u] = min(low[u],low[to]);
if(low[to]>=dfn[u]){
if(fa==-1)num_rt++;
else iscut[u] = 1;
while(1){
int tmp = sta.top();
sta.pop();
iscut[tmp] = 0;
instack[tmp] = 0;
lin[cnt].push_back(tmp);
if(tmp==to)break;
}
lin[cnt].push_back(u);
cnt++;
}
} else if(instack[to]) low[u] = min(low[u],dfn[to]);
}
}
int col
;
bool dfs(int id,int u,int c){
int i;
col[u] = c;
for(i=0;i<lin[id].size();i++){
int to = lin[id][i];
if(u==to || col[to]!=-1)continue;
if(g[u][to]){
dfs(id,to,c^1);
}
}
return true;
}
bool ok
;
int solve(int n){
int i,j;
for(i=1;i<=n;i++){
if(!dfn[i]){
num_rt = 0;
tarjan(i,-1);
if(num_rt>=2){
iscut[i] = 1;
}
}
}
int ans = 0;
memset(ok,false,sizeof(ok));
bool tag = 0;
for(i=0;i<cnt;i++){
for(j=0;j<=n;j++)col[j]=-1;
int sz = lin[i].size();
dfs(i,lin[i][0],1);
bool tag = 0;
for(j=0;j<sz;j++){
for(int k=0;k<sz;k++){
int a = lin[i][j],b = lin[i][k];
if(a!=b && g[a][b] && col[a]==col[b]){
tag = 1;
break;
}
}
if(tag)break;
}
if(tag){
for(j=0;j<sz;j++){
ok[lin[i][j]] = 1;
}
}
}
for(i=1;i<=n;i++){
if(!ok[i])ans++;
}
return ans;
}

int main(){
int n,m;
while(scanf("%d%d",&n,&m) &&(n+m)){
int i,j;
memset(g,true,sizeof(g));
init(n);
while(m--){
int x,y;
scanf("%d%d",&x,&y);
g[x][y] = g[y][x] = false;
}
for(i=0;i<=n;i++){
v[i].clear();
lin[i].clear();
}
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(g[i][j] && i!=j){
v[i].push_back(j);
}
}
}
printf("%d\n",solve(n));
}
return 0;
}