HDU 3081 Marriage Match II
题意：n个女孩n个男孩，每一个女孩能够和一些男孩配对。然后有些女孩是朋友。满足这个朋友圈里面的人，假设有一个能和某个男孩配对，其它就都能够，然后每轮要求每一个女孩匹配到一个男孩。且每轮匹配到的都不同，问最多能匹配几轮
思路：二分轮数k，然后建图为，源点连向女孩，男孩连向汇点容量都为k，然后女孩和男孩之间连边为。有关系的连边容量1，这样一个匹配相应一条边。且不会反复，每次推断最大流是否等于n * k就可以
代码：
#include#include #include #include using namespace std;const int MAXNODE = 205;const int MAXEDGE = 100005;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge { int u, v; Type cap, flow; Edge() {} Edge(int u, int v, Type cap, Type flow) { this->u = u; this->v = v; this->cap = cap; this->flow = flow; }};struct Dinic { int n, m, s, t; Edge edges[MAXEDGE]; int first[MAXNODE]; int next[MAXEDGE]; bool vis[MAXNODE]; Type d[MAXNODE]; int cur[MAXNODE]; vector cut; void init(int n) { this->n = n; memset(first, -1, sizeof(first)); m = 0; } void add_Edge(int u, int v, Type cap) { edges[m] = Edge(u, v, cap, 0); next[m] = first[u]; first[u] = m++; edges[m] = Edge(v, u, 0, 0); next[m] = first[v]; first[v] = m++; } bool bfs() { memset(vis, false, sizeof(vis)); queue Q; Q.push(s); d[s] = 0; vis[s] = true; while (!Q.empty()) { int u = Q.front(); Q.pop(); for (int i = first[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (!vis[e.v] && e.cap > e.flow) { vis[e.v] = true; d[e.v] = d[u] + 1; Q.push(e.v); } } } return vis[t]; } Type dfs(int u, Type a) { if (u == t || a == 0) return a; Type flow = 0, f; for (int &i = cur[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) { e.flow += f; edges[i^1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } Type Maxflow(int s, int t) { this->s = s; this->t = t; Type flow = 0; while (bfs()) { for (int i = 0; i < n; i++) cur[i] = first[i]; flow += dfs(s, INF); } return flow; } void MinCut() { cut.clear(); for (int i = 0; i < m; i += 2) { if (vis[edges[i].u] && !vis[edges[i].v]) cut.push_back(i); } }} gao;const int N = 105;int t, n, m, f, g[N][N], parent[N];int find(int x) { return x == parent[x] ? x : parent[x] = find(parent[x]);}bool judge(int k) { int s = 0, t = n * 2 + 1; gao.init(t + 1); for (int i = 1; i <= n; i++) { gao.add_Edge(s, i, k); gao.add_Edge(i + n, t, k); for (int j = 1; j <= n; j++) if (g[i][j]) gao.add_Edge(i, j + n, 1); } return gao.Maxflow(s, t) == n * k;}int main() { scanf("%d", &t); while (t--) { scanf("%d%d%d", &n, &m, &f); memset(g, 0, sizeof(g)); for (int i = 1; i <= n; i++) parent[i] = i; int u, v; while (m--) { scanf("%d%d", &u, &v); g[u][v] = 1; } while (f--) { scanf("%d%d", &u, &v); int pa = find(u); int pb = find(v); if (pa != pb) parent[pa] = pb; } for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { if (find(i) == find(j)) { for (int k = 1; k <= n; k++) g[i][k] |= g[j][k]; } } int l = 1, r = n + 1; while (l < r) { int mid = (l + r) / 2; if (judge(mid)) l = mid + 1; else r = mid; } printf("%d\n", l - 1); } return 0;}
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