染色法判断二分图

const int N = 1e5 + 10, M = 2 * N;int e[M], ne[M], h[N], n, m, idx = 0, color[N];
void add(int a, int b){e[idx] = b; ne[idx] = h[a]; h[a] = idx++;}bool dfs(int u, int c)
{color[u] = c;           // 染色该点for(int i = h[u]; i != -1; i = ne[i]){int j = e[i];if(color[j] == 0){                   // 染色其连通点，如果存在染色矛盾，立即返回falseif(dfs(j, 3 - c) == false)return false;}                   // 发生颜色矛盾，连通点的颜色相同else if(color[j] == c)return false;}return true;            // 对连通点的染色过程未发生矛盾
}int main()
{fill(h, h + N, -1);cin >> n >> m;while(m--){int a, b;cin >> a >> b;add(a, b);add(b, a);}bool flag = true;for(int i = 1; i <= n; ++i){if(color[i] == 0)flag = dfs(i, 1); if(flag == false)break;}if(flag)cout << "Yes";elsecout << "No";return 0;
}

匈牙利算法求二分图最大匹配

const int N = 510, M = 1e5 + 10;
int n1, n2, m, h[N], e[M], ne[M], idx = 0;
int match[N];
bool visited[N];
void add(int a, int b){e[idx] = b; ne[idx] = h[a]; h[a] = idx++;}int find(int u)
{for(int v = h[u]; v != -1; v = ne[v]){int j = e[v];if(visited[j] == false){                           // 若j已有匹配，试找与其匹配的点visited[j] = true;      // 是否还有其他相连通且未发生匹配的点if(match[j] == 0 || find(match[j])){match[j] = u;return true;}}}return false;
}
int main()
{fill(h, h + N, -1);cin >> n1 >> n2 >> m;while(m--){int a, b;cin >> a >> b;add(a, b);}int ret = 0;for(int i = 1; i <= n1; ++i){fill(visited, visited + N, false);  // 每次查询一个点的匹配，都要置空if(visited[i] == false)             // match保证了已匹配的点的状态{if(find(i))ret++;}}cout << ret;return 0;
}