hdu 1533 最小费用最大流模板题

这一题虽然说是模板题，但作为学习了最小费用最大流过程之后的第一题，模型的建立还是一脸懵比，之后通过见识相关的变形再来磨练建模的能力吧
如下是网上题解中关于本题中网络流的建模：
1.所有人到所有的房子均建容量为1，费用为人到房子的曼哈顿距离的流
2.建立超级源点s，s到所有人均建容量为1，费用为0的流
3.建立超级汇点t，所有房子到t均建容量为1，费用为0的流

#include
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 105*2;

int n,m;
char f[105][105];

struct node{
    int x,y;
    int dist(node b){
        return abs(x - b.x) + abs(y - b.y);
    }
}man[105],house[105];


struct Edge{
    int from, to, cap, flow, cost;
    Edge(int u, int v, int c, int f, int w): from(u),to(v),cap(c),flow(f),cost(w){}
};

struct MCMF{
    int n,m;
    vector edges;
    vector<int> G[maxn];
    int inq[maxn];
    int d[maxn];
    int p[maxn];
    int a[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, int cost){
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int s, int t, int& flow, long long& cost){
        for(int i = 0; i < n; ++i) d[i] = INF;
        memset(inq,0,sizeof(inq));
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;

        queue<int>Q;
        Q.push(s);
        while(!Q.empty()){
            int u = Q.front(); Q.pop(); inq[u] = 0;
            for(int i = 0; i < G[u].size(); ++i){
                Edge& e = edges[G[u][i]];
                if(e.cap > e.flow && d[e.to] > d[u] + e.cost){
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
                }
            }
        }
        if(d[t] == INF) return false;
        flow += a[t];
        cost += (long long)d[t] * (long long)a[t];
        for(int u = t; u != s; u = edges[p[u]].from){
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
        }
        return true;
    }

    int MincostMaxflow(int s, int t, long long& cost){
        int flow = 0; cost = 0;
        while(BellmanFord(s, t, flow, cost)) ;
        return flow;
    }
};

MCMF mc;

int main(){
    while(~scanf("%d%d",&n,&m) && (n || m)){
        memset(f,0,sizeof(f));
        memset(man,0,sizeof(man));
        memset(house,0,sizeof(house));
        for(int i = 1; i <= n; ++i){
            scanf("%s",f[i]+1);
        }
        int p = 0, q = 0;
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= m; ++j){
                if(f[i][j] == 'm'){
                    man[p].x = i;
                    man[p].y = j;
                    p++;
                }
                else if(f[i][j] == 'H'){
                    house[q].x = i;
                    house[q].y = j;
                    q++;
                }
            }
        }
        mc.init(p+q+2);
        for(int i = 0; i < p; ++i){
            for(int j = 0 ; j < q; ++j){
                mc.AddEdge(i,p+j,1,man[i].dist(house[j]));
            }
        }
        for(int i = 0; i < p; ++i)
            mc.AddEdge(p+q,i,1,0);
        for(int i = p; i < p+q; ++i)
            mc.AddEdge(i,p+q+1,1,0);
        long long ans;
        mc.MincostMaxflow(p+q,p+q+1,ans);
        printf("%lld\n",ans);
    }
    return 0;
}