hdu 2063 过山车（二分图最佳匹配）

　　经典的二分图最大匹配问题，因为匈牙利算法我还没有认真去看过，想先试试下网络流的做法，即对所有女生增加一个超级源，对所有男生增加一个超级汇，然后按照题意的匹配由女生向男生连一条边，跑一个最大流就是答案（以上所有边容量均为 1 ），我是直接上 Dinic 算法的模板的：

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
#define  sd(x)  scanf("%d",&(x))
const int inf = 0x3fffffff;

struct Edge {
    int from, to, cap, flow;
    Edge() {}
    Edge(int from, int to, int cap, int flow):
        from(from), to(to), cap(cap), flow(flow) {}
};

const int N = 1003;

struct Dinic {
    int s,t;
    vector<Edge> edges;
    vector<int> G[N];
    bool vis[N];
    int d[N];
    int cur[N];

    void clear() {
        edges.clear();
        for(int i = 0; i < N; ++i)
            G[i].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);
        d[s] = 0;
        vis[s] = 1;
        while(!q.empty()) {
            int x = q.front();  q.pop();
            int len = G[x].size();
            for(int i = 0; i < len; ++i) {
                Edge &e = edges[G[x][i]];
                if(!vis[e.to] && e.cap > e.flow) {
                    d[e.to] = d[x] + 1;
                    vis[e.to] = 1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int dfsAll(int x, int a) {
        if(x == t || a == 0)    return a;
        int flow = 0, f, len = G[x].size();
        for(int &i = cur[x]; i < len; ++i) {
            Edge &e = edges[G[x][i]];
            if(d[e.to] == d[x] + 1 && (f = dfsAll(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 += dfsAll(s,inf);
        }
        return flow;
    }
} dinic;

int c[1003];

#define sd2(x,y)  scanf("%d%d",&(x),&(y))
#define sd3(x,y,z)  scanf("%d%d%d",&(x),&(y),&(z))

int main() {
    int n,m,k,x,y;
    while(~sd3(k, m, n), k) {
        dinic.clear();
        while(k--) {
            sd2(x,y);
            y += m;
            dinic.addEdge(x, y, 1);
//            dinic.addEdge(y, x, 1);
        }
        for(int i = 1; i <= m; ++i)
            dinic.addEdge(0, i, 1);
        for(int i = m + 1; i <= m + n; ++i)
            dinic.addEdge(i, m + n + 1, 1);
        printf("%d
",dinic.Maxflow(0, m + n + 1));
    }
    return 0;
}

　　需要注意的是男女生的编号，还有女生指向男生的是有向边，不需要两次的 addEdge。wa 了两次才找出所有问题。。。