题意：L公司和H公司举办了一次联谊晚会。晚会上，L公司的N位员工和H公司的M位员工打算进行一场交际舞。在这些领导中，一些L公司的员工和H公司的员工之间是互相认识的，这样的认识关系一共有T对。舞会上，每位员工会尝试选择一名Ta认识的对方公司的员工作为舞伴，并且每位员工至多跳一支舞。完成的交际舞的数量越多，晚会的气氛就越热烈。顾及到晚会的气氛，员工们希望知道，哪些员工之间如果进行了交际舞，就会使整场晚会能够完成的交际舞的最大数量减小。
分析：求出二分图最大匹配的不可行边板子题，使用dinic求最大流，tarjan求强连通分量。详见《算法竞赛进阶指南》P446-447。
代码：
#include<bits/stdc++.h>
using namespace std;
const int inf = 0x3fffffff, u = 40010, w = 300010;
int head[u], ver[w], edge[w], Next[w], d[u], e[w], c[u], sta[u], ins[u], dfn[u], low[u];
int n, m, p, s, t, i, j, tot, maxflow, ans, x, y, scc, st, num;
char str[10];
vector<int> a[u];
queue<int> q;void add(int x, int y, int z) {ver[++tot] = y, edge[tot] = z, Next[tot] = head[x], head[x] = tot;ver[++tot] = x, edge[tot] = 0, Next[tot] = head[y], head[y] = tot;
}bool bfs() {memset(d, 0, sizeof(d));while (q.size()) q.pop();q.push(s); d[s] = 1;while (q.size()) {int x = q.front(); q.pop();for (int i = head[x]; i; i = Next[i])if (edge[i] && !d[ver[i]]) {q.push(ver[i]);d[ver[i]] = d[x] + 1;if (ver[i] == t) return 1;}}return 0;
}int dinic(int x, int flow) {if (x == t) return flow;int rest = flow, k;for (int i = head[x]; i && rest; i = Next[i])if (edge[i] && d[ver[i]] == d[x] + 1) {k = dinic(ver[i], min(rest, edge[i]));if (!k) d[ver[i]] = 0;edge[i] -= k;edge[i ^ 1] += k;rest -= k;}return flow - rest;
}void add2(int x, int y)
{a[x].push_back(y);
}void tarjan(int x)
{dfn[x] = ++num; low[x] = num;sta[++st] = x; ins[x] = 1;int y;for (int i = 0; i<a[x].size(); i++)if (!dfn[y = a[x][i]]){tarjan(y);low[x] = min(low[x], low[y]);}else if (ins[y]) low[x] = min(low[x], dfn[y]);if (dfn[x] == low[x]){scc++;do { y = sta[st]; st--; ins[y] = 0; c[y] = scc; } while (x != y);}
}int main()
{while (cin >> n >> m >> p){memset(head, 0, sizeof(head));s = 0, t = n + m + 1; tot = 1; maxflow = 0;for (i = 1; i <= n; i++) add(s, i, 1);for (i = 1; i <= m; i++) add(i + n, t, 1);for (i = 1; i <= p; i++){scanf("%d%d", &x, &y);add(x, n + y, 1), e[i] = tot;}while (bfs())while (i = dinic(s, inf)) maxflow += i;for (i = s; i <= t; i++) a[i].clear();for (i = 1; i <= p; i++)if (!edge[e[i]]) add2(ver[e[i]], ver[e[i] ^ 1]);else add2(ver[e[i] ^ 1], ver[e[i]]);for (i = 1; i <= n; i++)if (!edge[2 * i]) add2(i, s); else add2(s, i);for (i = 1; i <= m; i++)if (!edge[2 * (n + i)]) add2(t, n + i); else add2(n + i, t);memset(dfn, 0, sizeof(dfn));memset(ins, 0, sizeof(ins));memset(c, 0, sizeof(c));st = num = scc = ans = 0;for (i = s; i <= t; i++)if (!dfn[i]) tarjan(i);for (i = 1; i <= p; i++)if (edge[e[i]] || c[ver[e[i]]] == c[ver[e[i] ^ 1]]) ans++;cout << (ans = p - ans) << endl;if (!ans) cout << endl;for (i = 1; i <= p; i++)if (!edge[e[i]] && c[ver[e[i]]] != c[ver[e[i] ^ 1]])if (--ans) printf("%d ", i); else printf("%d\n", i);}return 0;
}