UVALive - 3268 Jamie's Contact Groups(二分＋最大流)

题目大意：有n个人和m个组，一个人可能属于多个组，现在请你从某些组中去掉几个人，使得每个人都只属于一个组，且人数最多的组中人员数目达到最小

解题思路：最大值最小化，二分
建图就比较简单了，二分主要二分组别到超级汇点的容量

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;#define N 2010#define INF 0x3f3f3f3fstruct Edge{    int from, to, cap, flow;    Edge() {}    Edge(int from, int to, int cap, int flow) : from(from), to(to), cap(cap), flow(flow) {}};struct Dinic{    int n, m, s, t;    vector<Edge> edges;    vector<int> G[N];    bool vis[N];    int d[N], cur[N];    void init(int n) {        this->n = n;        for (int i = 0; i <= n; i++) {            G[i].clear();        }        edges.clear();    }    void AddEdge(int from, int to, int cap) {        edges.push_back(Edge(from, to, cap, 0));        edges.push_back(Edge(to, from, 0, 0));        int m = edges.size();        G[from].push_back(m - 2);        G[to].push_back(m - 1);    }     bool BFS() {        memset(vis, 0, sizeof(vis));        queue<int> Q;        Q.push(s);        vis[s] = 1;        d[s] = 0;        while (!Q.empty()) {            int u = Q.front();            Q.pop();            for (int i = 0; i < G[u].size(); i++) {                Edge &e = edges[G[u][i]];                if (!vis[e.to] && e.cap > e.flow) {                    vis[e.to] = true;                    d[e.to] = d[u] + 1;                    Q.push(e.to);                }            }        }        return vis[t];    }    int DFS(int x, int a) {        if (x == t || a == 0)            return a;        int flow = 0, f;        for (int i = cur[x]; i < G[x].size(); i++) {            Edge &e = edges[G[x][i]];            if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {                e.flow += f;                edges[G[x][i] ^ 1].flow -= f;                flow += f;                a -= f;                if (a == 0)                    break;            }        }        return flow;    }    int Maxflow(int s, int t) {        this->s = s; this->t = t;        int flow = 0;        while (BFS()) {            memset(cur, 0, sizeof(cur));            flow += DFS(s, INF);        }        return flow;    }    bool MinCut(int mid, int m, int n, int source, int sink) {        for (int i = 0; i < edges.size(); i++)            edges[i].flow = 0;        for (int i = 0; i < m; i++)             for (int j = 0; j < G[i].size(); j++)                 if (edges[G[i][j]].to == sink)                    edges[G[i][j]].cap = mid;        int t = Maxflow(source, sink);        if (t == n) {            return true;        }        return false;    }};Dinic dinic;int n, m, source, sink;int num[N], Max;char name[30]; void init() {    source = n + m; sink = source + 1;    memset(num, 0, sizeof(num));    dinic.init(sink);    Max = -INF;    int t;    char c;    for (int i = 0; i < n; i++) {        dinic.AddEdge(source, m + i, 1);        scanf("%s", name);        c = getchar();        while (c != '\n') {            scanf("%d", &t);            dinic.AddEdge(m + i, t, 1);            num[t]++;            c = getchar();        }    }    for (int i = 0; i < m; i++) {        Max = max(Max, num[i]);        dinic.AddEdge(i, sink, num[i]);    }}void solve() {//  printf("Max is %d\n", Max);    int l = 0, r = Max, mid, ans = 0;    while (l <= r) {        mid = (l + r) / 2;        if (dinic.MinCut(mid, m, n, source, sink)) {            r = mid - 1;            ans = mid;        }        else l = mid + 1;     }    printf("%d\n", ans);}int main() {    while (scanf("%d%d", &n, &m) != EOF && n + m) {        init();        solve();    }    return 0;}