Going Home

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man. 

Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a ‘.‘ means an empty space, an ‘H‘ represents a house on that point, and am ‘m‘ indicates there is a little man on that point. 

You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

2
10
28

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
typedef long long ll;
using namespace std;
/*最小费用最大流，求最大费用只需要取相反数，结果取相反数即可。
点的总数为N，点的编号0~N-1*/
const int MAXN = 300;
const int MAXM = 100000;

struct Node
{
    int x,y;
}H[MAXN],M[MAXN];

struct Edge
{
    int to,next,cap,flow,cost;
}edge[MAXM];

int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N,n,m;

void init(int n)
{
    N=n;
    tol=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int cap,int cost)
{
    edge[tol].to=v;
    edge[tol].cap=cap;
    edge[tol].cost=cost;
    edge[tol].flow=0;
    edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].to=u;
    edge[tol].cap=0;
    edge[tol].cost=-cost;
    edge[tol].flow=0;
    edge[tol].next=head[v];
    head[v]=tol++;
}

bool spfa(int s,int t)
{
    queue<int>q;
    for (int i=0;i<N;i++)
    {
        dis[i]=INF;
        vis[i]=false;
        pre[i]=-1;
    }
    dis[s]=0;
    vis[s]=true;
    q.push(s);
    while (!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=false;
        for (int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost)
            {
                dis[v]=dis[u] + edge[i].cost;
                pre[v]=i;
                if (!vis[v])
                {
                    vis[v]=true;
                    q.push(v);
                }
            }
        }
    }
    if (pre[t]==-1) return false;
    else return true;
}

//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
    int flow=0;
    cost=0;
    while (spfa(s,t))
    {
        int Min=INF;
        for (int i=pre[t];i!=-1;i=pre[edge[i^1].to])
        {
            if (Min > edge[i].cap-edge[i].flow)
                Min=edge[i].cap-edge[i].flow;
        }
        for (int i=pre[t];i!=-1;i=pre[edge[i^1].to])
        {
            edge[i].flow+=Min;
            edge[i^1].flow-=Min;
            cost+=edge[i].cost*Min;
        }
        flow+=Min;
    }
    return flow;
}

int main()
{
    char ch;
    while (scanf("%d%d",&n,&m)&&(m||n))
    {
        int hh=1,mm=1;
        getchar();
        for (int i=1;i<=n;i++)
        {
            for (int j=1;j<=m;j++)
            {
                scanf("%c",&ch);
                if (ch=='H')
                {
                    H[hh].x=i;
                    H[hh++].y=j;
                }
                else if (ch=='m')
                {
                    M[mm].x=i;
                    M[mm++].y=j;
                }
            }
            getchar();
        }
        N=hh+mm+2;
        init(N);
        mm--,hh--;
        for (int i=1;i<=mm;i++)
            addedge(0,i,1,0);
        for (int i=1;i<=hh;i++)
            addedge(i+mm,N-1,1,0);
        for (int i=1;i<=mm;i++)
        {
            for (int j=1;j<=hh;j++)
            {
                int x=abs(M[i].x-H[j].x)+abs(M[i].y-H[j].y);
                addedge(i,j+mm,1,x);
            }
        }
        int ans;
        int y=minCostMaxflow(0,N-1,ans);
        printf("%d\n",ans);
    }
    return 0;
}

Going Home (hdu 1533 最小费用流)

原文：http://blog.csdn.net/u014422052/article/details/45726541