UVA 1345 - Jamie's Contact Groups（二分+最大流）

UVA 1345 - Jamie's Contact Groups

题目链接

题意：给定一些人，每个人有一个分组，现在要每个人选一个分组，使得所有分组中最大的人数最小，问这个最小值是多少

思路：二分答案，然后利用最大流去判定，源点往每个人建一条边容量为1，每个人往各自的分组建一条边，容量为1，分组向汇点建一条边，容量为二分出来的值，这样跑一下最大流如果最大流等于n，就是能满足

代码：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#include <map>#include <string>#include <iostream>#include <set>using namespace std;const int MAXNODE = 1505;const int MAXEDGE = 1100005;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<int> 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<int> 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;}bool Maxflow(int s, int t, int tot) {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 == tot;}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 = 1005;const int M = 505;int n, m, cnt[M];vector<int> f[N];char str[100005];bool judge(int c) {gao.init(n + m + 2);int s = 0, t = n + m + 1;for (int i = 0; i < n; i++) {gao.add_Edge(0, i + 1, 1);for (int j = 0; j < f[i].size(); j++) {gao.add_Edge(i + 1, f[i][j] + n + 1, 1);}}for (int i = 0; i < m; i++)gao.add_Edge(n + i + 1, t, c);return gao.Maxflow(s, t, n);}int main() {while (~scanf("%d%d", &n, &m) && n || m) {memset(cnt, 0, sizeof(cnt));while ((getchar()) != '\n');for (int i = 0; i < n; i++) {f[i].clear();gets(str);int len = strlen(str);int sum = 0;int flag = 0;for (int j = 0; j < len; j++) {if (str[j] >= '0' && str[j] <= '9') {flag = 1;sum = sum * 10 + str[j] - '0';} else {if (flag) {f[i].push_back(sum);cnt[sum]++;}flag = 0;sum = 0;}}if (flag) {f[i].push_back(sum);cnt[sum]++;}}int l = 0, r = 0;for (int i = 0; i < m; i++) r = max(r, cnt[i]);while (l < r) {int mid = (l + r) / 2;if (judge(mid)) r = mid;else l = mid + 1;}printf("%d\n", l);}return 0;}