题意：找出图中不可能在奇圈中的点. [分析]注意到，在不同点双连通分量中的两个点，显然是不会存在圈的.那么这样，问题就划归为在点双连通分量中去找奇圈。 [重要性质]在一个点双连通分量中,只要有任意一个奇圈，那么所有的点都可以在一个奇圈内(证明看《算法竞赛入门经典 训练指南》). [重要定理]一个图含奇圈当且仅当图不是二分图. [解题思路]先求出图的点双连通分量(块),然后对每一个块染色判断二分图,统计出不可能在奇圈中的点的个数 [注意]染色判定二分图的算法要写对

#include
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#include
#define MID(x,y) ((x+y)>>1)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;const int N = 1002;
const int E = 2000002;
struct node{int u, v;int next;
}arc[E];
int cnt, head[N];
void init(){mem(head, -1);cnt = 0;
}
void add(int u, int v){arc[cnt].u = u;arc[cnt].v = v;arc[cnt].next = head[u];head[u] = cnt ++;arc[cnt].u = v;arc[cnt].v = u;arc[cnt].next = head[v];head[v] = cnt ++;
}/* 求点双连通分量 */
int dfn[N], low[N];
set  bcc[N];
int id, bcc_num;
stack  st;
void addbcc(int u, int v){bcc[bcc_num].insert(u);bcc[bcc_num].insert(v);
}
void dfs(int u, int father){dfn[u] = low[u] = ++id;for (int i = head[u]; i != -1; i = arc[i].next){int v = arc[i].v;if (v == father)    continue;if (dfn[v] < dfn[u]){st.push(arc[i]);if (!dfn[v]){dfs(v, u);low[u] = min(low[u], low[v]);if (dfn[u] <= low[v]){++ bcc_num;while(!st.empty()){int a = st.top().u;int b = st.top().v;st.pop();addbcc(a, b);if ((a == u && b == v) || (b == u && a == v) )break;}}}else{low[u] = min(low[u], dfn[v]);}}}
}
void bcc_tarjan(int n){id = bcc_num = 0;mem(dfn, 0);mem(low, 0);while(!st.empty())st.pop();for (int i = 0; i <= n; i ++)bcc[i].clear();for (int i = 1; i <= n; i ++)dfs(i, 0);
}
/* 求点双连通分量 *//* 染色判定二分图 */
int col[N];
bool not_bigragh[N];    //标记某个点双连通分量是不是二分图
void dfs_color(int bcc_id, int u, int color){col[u] = color;for (int i = head[u]; i != -1; i = arc[i].next){if (not_bigragh[bcc_id])return;int v = arc[i].v;if (bcc[bcc_id].find(v) == bcc[bcc_id].end())continue;if (col[v] == col[u]){not_bigragh[bcc_id] = 1;return;}else if (col[v] == -1){dfs_color(bcc_id, v, (color+1)&1);}}
}
bool fill(int bcc_id){      //对某个点双连通分量染色判断二分图set  ::iterator it;for (it = bcc[bcc_id].begin(); it != bcc[bcc_id].end(); it ++){not_bigragh[bcc_id] = 0;int u = *it;mem(col, -1);dfs_color(bcc_id, u, 0);if (not_bigragh[bcc_id])return false;}return true;
}
/* 染色判定二分图 */int res;
bool can[N];        //存某个点能否在一个奇圈中
int solve(int n){res = 0;mem(can, 0);for (int i = 1; i <= bcc_num; i ++){if (!fill(i)){set  ::iterator it;for (it = bcc[i].begin(); it != bcc[i].end(); it ++){can[*it] = 1;}}}for (int i = 1; i <= n; i ++){if (!can[i])res ++;}return res;
}
bool mat[N][N];     //表示i憎恨j的关系矩阵
int main(){int n, m;while(scanf("%d %d", &n, &m) != EOF){if (n == 0 && m == 0)break;init();mem(mat, 0);for (int i = 0; i < m; i ++){int a, b;scanf("%d %d", &a, &b);mat[a][b] = 1;mat[b][a] = 1;}for (int i = 1; i <= n; i ++){for (int j = i+1; j <= n; j ++){if (!mat[i][j]){add(i, j);}}}bcc_tarjan(n);printf("%d\n", solve(n));}return 0;
}