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http://acm.hdu.edu.cn/showproblem.php?pid=1533

C++版本一

题解：最小费用最大流

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200+10;
const int M=100+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int s,t,n,m,k,p,l,r,u,v;
int ans,cnth,cntm,flag,temp,sum,minz,maxflow;
int dis[N];
bool vis[N];
char str[N][N];
struct Node{
    int x,y;
    Node(){};
    Node(int X,int Y):x(X),y(Y){}
}home[N],man[N];
struct node{
    int u,v,c,w;
    node(){};
    node(int form,int to,int cap,int w):u(form),v(to),c(cap),w(w){}
};
vector<node>edge;
vector<int> G[N];
void Addedge(int u,int v,int cap,int w){
    edge.push_back({u,v,cap,w});
    edge.push_back({v,u,0,-w});
    int sz=edge.size();
    G[u].push_back(sz-2);
    G[v].push_back(sz-1);
}
bool bfs(int u){
    //memset(dis,-1,sizeof(dis));
    for(int i=0;i<=t;i++)dis[i]=INF,vis[i]=0;
    dis[u]=0;
    queue<int>q;
    q.push(u);
    vis[u]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=0;i<G[u].size();i++){
            node e=edge[G[u][i]];//cout<<u<<" "<<e.v<<endl;
            if(dis[e.v]>dis[u]+e.w&&e.c>0){
                dis[e.v]=dis[u]+e.w;
                if(!vis[e.v]){
                    vis[e.v]=1;
                    q.push(e.v);
                }
            }
        }
    }
    return dis[t]!=INF;
}
int dfs(int u,int flow){
    vis[u]=1;
    if(u==t)
        return flow;
    int now;
    for(int i=0;i<G[u].size();i++){
        node e=edge[G[u][i]];//cout<<ans<<endl;
        if((!vis[e.v]||e.v==t)&&e.c>0&&dis[u]+e.w==dis[e.v]&&(now=dfs(e.v,min(flow,e.c)))){
            ans+=e.w*now;
            //cout<<now<<endl;
            edge[G[u][i]].c-=now;
            edge[G[u][i]^1].c+=now;
            return now;
        }
    }
    return 0;
}
void dinic(){
    while(bfs(s)){
        int res=0;
        //memset(vis,0,sizeof(vis));
        while((res=dfs(s,INF))){
            maxflow+=res;
            memset(vis,0,sizeof(vis));
        }
    }
}
void init(){
    for(int i=0;i<=t;i++)G[i].clear();
    edge.clear();
    ans=0;
    maxflow=0;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    while(~scanf("%d%d",&n,&m)&&n+m){
        cnth=cntm=0;
        for(int i=1;i<=n;i++){
            scanf("%s",str[i]+1);
            for(int j=1;j<=m;j++){
                if(str[i][j]=='H')home[++cnth]=Node(i,j);
                if(str[i][j]=='m')man[++cntm]=Node(i,j);
            }
        }
        s=0;
        t=cnth+cntm+1;
        init();
        for(int i = 1; i <= cnth; i++){//建图
           for(int j = 1; j <= cntm; j++){
                int dist= abs(home[i].x - man[j].x) + abs(home[i].y - man[j].y);
                Addedge(i,j+cnth,1,dist);
                //Addedge(j+cnth,i,0,-dist);
            }
        }
        for(int i=1;i<=cnth;i++){
            Addedge(s,i,1,0);
            //Addedge(i,s,0,0);
        }
        for(int i=cnth+1;i<t;i++){
            //Addedge(t,i,0,0);
            Addedge(i,t,1,0);
        }
        //cout<<s<<" "<<t<<endl;
        dinic();
        cout<<ans<<endl;
    }

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

题解：带权二分图的最优匹配

每个man和house建立带权二分图，曼哈顿距离就是边的值，这里要求最小费用，也就是二分图最小权值匹配，但是KM算法求的是二分图最大权值匹配，所以此处用边的负数求最优匹配，求出来的答案的负数就是最小权匹配。

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