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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>

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