poj2942 Knights of the Round Table，无向图点双联通，二分图判定

<span style="font-size:18px;">#include <cstdio>
#include <stack>
#include <vector>
#include <algorithm>
#include <cstring>

using namespace std;

struct Edge{
int u, v;
};
const int maxn = 1005;
int pre[maxn], iscut[maxn], bccno[maxn],dfs_clock, bcc_cnt;
vector<int> G[maxn], bcc[maxn];

stack<Edge> S;

int dfs(int u, int fa){
int lowu = pre[u] = ++dfs_clock;
int child = 0;
for(int i=0; i<G[u].size(); ++i){
int v = G[u][i];
Edge e = (Edge){u, v};
if(!pre[v]){
S.push(e);
child++;
int lowv = dfs(v, u);
lowu = min(lowu, lowv);
if(lowv>=pre[u]){
iscut[u] = true;
bcc_cnt++; bcc[bcc_cnt].clear();
for(;;){
Edge x = S.top(); S.pop();
if(bccno[x.u]!=bcc_cnt){
bcc[bcc_cnt].push_back(x.u); bccno[x.u] = bcc_cnt;
}
if(bccno[x.v] != bcc_cnt){
bcc[bcc_cnt].push_back(x.v); bccno[x.v] = bcc_cnt;
}
if(x.u==u && x.v==v) break;
}
}
}
else if(pre[v]<pre[u] &&v!=fa){
S.push(e);
lowu = min(lowu, pre[v]);
}
}
if(fa < 0 && child == 1) iscut[u] = 0;
return lowu;
}

void find_bcc(int n){
memset(pre, 0, sizeof pre );
memset(iscut, 0, sizeof iscut );
memset(bccno, 0, sizeof bccno );
dfs_clock = bcc_cnt = 0;
for(int i=0; i<n; ++i)
if(!pre[i]) dfs(i, -1);
}

int odd[maxn], color[maxn];
bool bipartite(int u, int b){
for(int i=0; i<G[u].size(); ++i){
int v = G[u][i]; if(bccno[v]!=b)continue;
if(color[v]==color[u]) return false;
if(!color[v]){
color[v] = 3 - color[u];
if(!bipartite(v, b)) return false;
}
}
return true;
}

int A[maxn][maxn];

int main(){
#ifndef ONLINE_JUDGE
freopen("in.cpp", "r", stdin);
freopen("out.cpp", "w", stdout);
#endif // ONLINE_JUDGE
int kase = 0, n, m;
while(scanf("%d%d", &n, &m)==2 &&n){
for(int i=0; i<n; ++i)G[i].clear();
memset(A, 0, sizeof A );
for(int i=0; i<m; ++i){
int u, v;
scanf("%d%d", &u, &v);
u--; v--;
A[u][v] = A[v][u] = 1;
}
for(int u=0; u<n; ++u)//避免重边
for(int v=u+1; v<n; ++v)
if(!A[u][v]) {
G[u].push_back(v);
G[v].push_back(u);
}

find_bcc(n);

memset(odd, 0, sizeof odd );
for(int i=1; i<=bcc_cnt; ++i){
memset(color, 0, sizeof color );
for(int j=0; j<bcc[i].size(); ++j) bccno[bcc[i][j]] = i;
int u = bcc[i][0];
color[u] = 1;
if(!bipartite(u, i))
for(int j=0; j<bcc[i].size(); ++j) odd[bcc[i][j]] = 1;
}
int ans = n;
for(int i=0; i<n; ++i) if(odd[i]) ans--;
printf("%d\n", ans);
}
return 0;
}
</span>