poj2289 Jamie's Contact Groups

思路：

二分+最大流。
实现：

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <string.h>
#include <assert.h>
#include <queue>
#include <vector>
#include <algorithm>
#include <iostream>
#include <sstream>

#define N (1500 + 2)
#define M (N * N + 4 * N)

typedef long long LL;

using namespace std;

struct edge 
{
    int v, cap, next;
};
edge e[M];

int head[N], level[N], cur[N];
int num_of_edges;

/*
 * When there are multiple test sets, you need to re-initialize before each
 */
void dinic_init(void) 
{
    num_of_edges = 0;
    memset(head, -1, sizeof(head));
    return;
}

int add_edge(int u, int v, int c1, int c2) 
{
    int& i = num_of_edges;

    assert(c1 >= 0 && c2 >= 0 && c1 + c2 >= 0); // check for possibility of overflow
    e[i].v = v;
    e[i].cap = c1;
    e[i].next = head[u];
    head[u] = i++;

    e[i].v = u;
    e[i].cap = c2;
    e[i].next = head[v];
    head[v] = i++;
    return i;
}

void print_graph(int n) 
{
    for (int u = 0; u < n; u++) 
    {
        printf("%d: ", u);
        for (int i = head[u]; i >= 0; i = e[i].next) 
        {
            printf("%d(%d)", e[i].v, e[i].cap);
        }
        printf("
");
    }
    return;
}

/*
 * Find all augmentation paths in the current level graph
 * This is the recursive version
 */
int dfs(int u, int t, int bn) 
{
    if (u == t) return bn;
    int left = bn;
    for (int &i = cur[u]; i >= 0; i = e[i].next) 
    {
        int v = e[i].v;
        int c = e[i].cap;
        if (c > 0 && level[u] + 1 == level[v]) 
        {
            int flow = dfs(v, t, min(left, c));
            if (flow > 0) 
            {
                e[i].cap -= flow;
                e[i ^ 1].cap += flow;
                cur[u] = i;
                left -= flow;
                if (!left) break;
            }
        }
    }
    if (left > 0) level[u] = 0;
    return bn - left;
}

bool bfs(int s, int t) 
{
    memset(level, 0, sizeof(level));
    level[s] = 1;
    queue<int> q;
    q.push(s);
    while (!q.empty()) 
    {
        int u = q.front();
        q.pop();
        if (u == t) return true;
        for (int i = head[u]; i >= 0; i = e[i].next) 
        {
            int v = e[i].v;
            if (!level[v] && e[i].cap > 0) 
            {
                level[v] = level[u] + 1;
                q.push(v);
            }
        }
    }
    return false;
}

LL dinic(int s, int t) 
{
    LL max_flow = 0;

    while (bfs(s, t)) 
    {
        memcpy(cur, head, sizeof(head));
        max_flow += dfs(s, t, INT_MAX);
    }
    return max_flow;
}

vector<int> v[N];
int n, m;
bool check(int x)
{
    dinic_init();
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j < v[i].size(); j++)
        {
            add_edge(i, v[i][j] + n + 1, 1, 0);
        }
    }
    for (int i = 1; i <= n; i++)
        add_edge(0, i, 1, 0);
    for (int j = n + 1; j <= n + m; j++)
    {
        add_edge(j, n + m + 1, x, 0);
    }
    return dinic(0, n + m + 1) == n;
}

int main() 
{
    while (cin >> n >> m, n || m)
    {
        getchar();
        string s, name;
        int group;
        for (int i = 1; i <= n; i++) v[i].clear();
        for (int i = 1; i <= n; i++)
        {
            getline(cin, s);
            stringstream ss(s);
            ss >> name;
            while (ss >> group)
            {
                v[i].push_back(group);
            }
        }
        int l = 0, r = n, ans = n;
        while (l <= r)
        {
            int m = (l + r) >> 1;
            if (check(m))
            {
                r = m - 1; ans = m;
            }
            else l = m + 1;
        }
        cout << ans << endl;
    }
    return 0;
}