分房间（最大流）

题目描述
给定\(n\)个人和\(m\)个房间，每个房间最多只能住一个人。

再给定\(g[i][j]\)表示第\(i\)个人是否愿意住在第\(j\)个房间。

问最多能住下多少人。
输入
第一行包含一个正整数\(T(1\leq T\leq 10)\)，表示测试数据的组数。

每组数据第一行包含两个正整数\(n,m(1\leq n,m\leq 50)\)。

接下来\(n\)行，每行一个长度为\(m\)的01串，如果第\(i\)行第\(j\)个字符为'1'，说明第\(i\)个人愿意住在第\(j\)个房间，否则说明不愿意。
输出
对于每组数据输出一行一个整数，即最多能住下的人数。

简单的最大流使用ISAP板子

#include<bits/stdc++.h>
using namespace std;
typedef vector<string> VS;
const int N = 110,INF = 1e9;
struct Edge {
  int from, to, cap, flow;
  Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};

bool operator < (const Edge& a, const Edge& b) {
  return a.from < b.from || (a.from == b.from && a.to < b.to);
}

struct ISAP {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[N];
  bool vis[N];
  int d[N],cur[N],p[N],num[N];

  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(t);
    vis[t] = 1;
    d[t] = 0;
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i] ^ 1];
        if (!vis[e.from] && e.cap > e.flow) {
          vis[e.from] = 1;
          d[e.from] = d[x] + 1;
          Q.push(e.from);
        }
      }
    }
    return vis[s];
  }

  void init(int n) {
    this->n = n;
    for (int i = 0; i < n; i++) G[i].clear();
    edges.clear();
  }

  int Augment() {
    int x = t, a = INF;
    while (x != s) {
      Edge& e = edges[p[x]];
      a = min(a, e.cap - e.flow);
      x = edges[p[x]].from;
    }
    x = t;
    while (x != s) {
      edges[p[x]].flow += a;
      edges[p[x] ^ 1].flow -= a;
      x = edges[p[x]].from;
    }
    return a;
  }

  int Maxflow(int s, int t) {
    this->s = s;
    this->t = t;
    int flow = 0;
    BFS();
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; i++) num[d[i]]++;
    int x = s;
    memset(cur, 0, sizeof(cur));
    while (d[s] < n) {
      if (x == t) {
        flow += Augment();
        x = s;
      }
      int ok = 0;
      for (int i = cur[x]; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if (e.cap > e.flow && d[x] == d[e.to] + 1) {
          ok = 1;
          p[e.to] = G[x][i];
          cur[x] = i;
          x = e.to;
          break;
        }
      }
      if (!ok) {
        int m = n - 1;
        for (int i = 0; i < G[x].size(); i++) {
          Edge& e = edges[G[x][i]];
          if (e.cap > e.flow) m = min(m, d[e.to]);
        }
        if (--num[d[x]] == 0) break;
        num[d[x] = m + 1]++;
        cur[x] = 0;
        if (x != s) x = edges[p[x]].from;
      }
    }
    return flow;
  }
};
ISAP FG;
void solve()
{
	int n,m;
	cin>>n>>m;
	VS g(n);
	for(int i=0;i<n;++i) cin>>g[i];
	int s = 0,t = n+m+1;
	//注意初始化
	FG.init(n+m+2);
	//从源点向所有的人连一条权值为1的边表示该人有房间或无房间
	for(int i=1;i<=n;++i) FG.AddEdge(s,i,1);
	//若人和房间匹配则加一条权值为1的边
	for(int i=0;i<n;++i) 
	 for(int j=0;j<m;++j)
	 if(g[i][j]=='1') FG.AddEdge(i+1,j+1+n,1);
	//从房间向汇点连一条权值为一的边表示该房间有人住
	for(int i=1;i<=m;++i) FG.AddEdge(n+i,t,1);
	//一条S->T的路径表示一个人选择了一个房间
	cout<<FG.Maxflow(s,t)<<'\n';
}
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		solve();
	}
}

使用hopcroft复杂度\(\sqrt{n}m\)

#include<bits/stdc++.h>
using namespace std;
typedef vector<string> VS;
const int N = 110;
const int INF = 1e9;
struct Hopcroft_Carp {
    int nx, ny, e, mx[N], my[N], dx[N], dy[N], dis;
    bool used[N];
    std::vector<int> G[N];
 
    void init(int nx,int ny)
    {
        this->nx = nx;
        this->ny = ny;
        for(int i = 1; i<=nx; ++i) G[i].clear();
        memset(mx,0,sizeof mx);
        memset(my,0,sizeof my);
    }
    void AddEdge(int u,int v)
    {
        G[u].push_back(v);
    }
    bool BFS() {
        std::queue<int> Q;
        dis = INF;
        memset(dx, -1, sizeof dx);
        memset(dy, -1, sizeof dy);
        for (int i = 1; i <= nx; ++ i) {
             if (mx[i] == 0) Q.push(i), dx[i] = 0;
        }
        while (!Q.empty()) 
        {
            int  u = Q.front(); Q.pop();
            if (dx[u] > dis) break;
            int size = G[u].size();
            for (int i = 0; i < size; ++ i) 
            {
                int v = G[u][i];
                if (dy[v] == -1) 
                {
                    dy[v] = dx[u] + 1;
                    if (my[v] == 0)
                        dis = dy[v];
                     else
                        dx[my[v]] = dy[v] + 1, Q.push(my[v]);
                }   
            }
         }
        return dis != INF;
    }
    
    bool DFS(int u) {
        int size = G[u].size();
        for (int i = 0; i < size; ++ i) {
            int v = G[u][i];
           if (!used[v] && dy[v] == dx[u] + 1) {
                used[v] = 1;
              if (my[v] != 0 && dy[v] == dis) continue;
                if (my[v] == 0 || DFS(my[v])) {
                    my[v] = u, mx[u] = v;
                   return 1;
                 }
             }
         }
        return 0;
     }
     
     int Maxmatch() {
        int ans = 0;
        while (BFS()) {
            memset(used, 0, sizeof used);
             for (int i = 1; i <= nx; ++ i)
                 if (mx[i] == 0 && DFS(i)) ++ ans;
        }
        return ans;
    }
}HC;
void solve()
{
    int n,m;
    cin>>n>>m;
    HC.init(n,m);
    VS g(n);
    for(int i=0;i<n;++i) cin>>g[i];
    for(int i=0;i<n;++i)
     for(int j=0;j<m;++j)
     if(g[i][j]=='1') HC.AddEdge(i+1,j+1);
    cout<<HC.Maxmatch()<<'\n';
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        solve();
    }
}