（二分/网络流）Optimal Milking

http://poj.org/problem?id=2112

题意为n个奶牛站，m头奶牛，每个奶牛站的奶牛上限k；给出奶牛站和奶牛对其他实体的距离，距离为0即为两个实体无路径，求出所有奶牛到奶牛站中走的最远的奶牛的最远距离的最小值
最大中的最小问题，可以二分。我们先用Floyd处理出实体到实体之间的最近距离，二分这个最远距离，建图（两实体距离小于等于二分值时才连边）后跑出最大流量判断是否为奶牛的数量

#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cstdio>
#include <queue>
using namespace std;
const int maxn = 300;
const int inf = 0x3f3f3f3f;
int path[maxn][maxn];
struct Edge {
    int from, to, cap, flow;
    Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};
struct Dinic{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> g[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    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));
        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);
        d[s] = 0;
        vis[s] = 1;
        while(!q.empty()){
            int x = q.front(); q.pop();
            for(int i = 0; i < (int)g[x].size(); i++){
                Edge &e = edges[g[x][i]];
                if(!vis[e.to] && e.cap > e.flow){
                    vis[e.to] = 1;
                    d[e.to] = d[x] + 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 < (int)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;
    }
} dinic;
int n, m, k;
void floyd() {
    for(int q = 1; q <= n + m; ++q) {
        for(int i = 1; i <= n + m; ++i) {
            if(path[i][q] == inf) continue;
            for(int j = 1; j <= n + m; ++j) {
                if(path[q][j] == inf) continue;
                path[i][j] = min(path[i][j], path[i][q] + path[q][j]);
            }
        }
    }
}
bool solve(int x) {
    dinic.init(n + m + 2);
    for(int i = 1; i <= n; ++i) {
        dinic.AddEdge(0, i, k);
        for(int j = n + 1; j <= n + m; ++j) {
            if(path[i][j] <= x) dinic.AddEdge(i, j, 1);
        }
    }
    for(int i = n + 1; i <= n + m; ++i) dinic.AddEdge(i, n + m + 1, 1);
    return dinic.Maxflow(0, n + m + 1) == m;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    while(cin >> n >> m >> k) {
        for(int i = 1; i <= n + m; ++i) {
            for(int j = 1; j <= n + m; ++j) {
                cin >> path[i][j];
                if(path[i][j] == 0) path[i][j] = inf;
            }
        }
        floyd();
        int l = 0, r = 1000000;
        while(l < r) {
            int mid = l + r >> 1;
            if(solve(mid)) r = mid;
            else l = mid + 1;
        }
        cout << l << endl;
    }
}