POJ 2195 Going Home 费用流 最小费用流

题目连接：http://poj.org/problem?id=2195

题目大意 ：题目的意思就是给你一个map然后m代表人，h代表house，一个人只能进入一个house，一个house只能住一个人，问你所以人进入house所能用的最小步数是多少~

思路：把man和house分别看做一个集合，这样的话你就可以对每个人和每个house 连接，cao = 1,cost = abs(x1-x2)+abs(y1-y2);然后开一个超级源点就可以了。其实模型就是走多少次就是容量。费用根据题目而定。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
const int maxn = 10050;
const int inf = 100000;
char str[105][105];
struct node
{
    int x,y;
};
struct edge
{
    int u,v,cap,flow,cost;
};
vector<struct node>house;
vector<struct node>man;
vector<int>g[10050];
vector<struct edge>e;
void init(int n)
{
    for(int i = 0;i <= n;i++)
    g[i].clear();
    e.clear();
    return;
}
void addedge(int u,int v,int cap,int flow,int cost)
{
    e.push_back((struct edge){u,v,cap,flow,cost});
    e.push_back((struct edge){v,u,0,flow,-cost});
    int m;
    m = e.size();
    g[u].push_back(m-2);
    g[v].push_back(m-1);
    return ;
}
int vis[maxn];
int pre[maxn];
int dis[maxn];
int a[maxn];
int spfa(int s,int t,int n,int &flow,int &cost)
{
    int i,u,v;
    for(i = 0;i <= n;i++)
    dis[i] = inf;
    memset(vis,0,sizeof(vis));
    dis[s] = pre[s] = 0;
    a[s] = inf;
    queue<int>q;
    q.push(s);
    vis[s] = 1;
    while(!q.empty())
    {

        u = q.front();
        vis[u] = 0;
        q.pop();

        for(i = 0;i < g[u].size();i++)
        {
            struct edge & ei = e[g[u][i]];
            v = ei.v;
            if(ei.cap > ei.flow && dis[v] > dis[u]+ei.cost)
            {
                dis[v] = dis[u]+ei.cost;
                pre[v] = g[u][i];
                a[v] = min(a[u],ei.cap-ei.flow);
                if(!vis[v])
                q.push(v),vis[v] = 1;
            }
        }
    }
    if(dis[t] == inf)
    return 0;

    flow += a[t];
    cost += a[t]*dis[t];

    u = t;
    while(u != s)
    {
        e[pre[u]].flow += a[t];
        e[pre[u]^1].flow -= a[t];
        u = e[pre[u]].u;
    }
    return 1;
}

int mcmf(int s,int t,int n)
{
    int flow = 0;
    int cost = 0;
    while(spfa(s,t,n,flow,cost));
    return cost;
}
int main()
{
    int m,n,i,j;
    while(scanf("%d %d",&n,&m)&&(m||n))
    {
        for(i = 0;i < n;i++)
        scanf("%s",str[i]);
        house.clear();
        man.clear();
        for(i = 0;i < n;i++)
        {
            for(j = 0;j < m;j++)
            {
                if(str[i][j] == 'H')
                house.push_back((struct node){i,j});
                else if(str[i][j] == 'm')
                man.push_back((struct node){i,j});
            }
        }

        int hsize,msize;
        hsize = house.size();
        msize = man.size();

        init(hsize+msize+5);
        for(i = 0;i < msize;i++)
        {
            for(j = 0;j < hsize;j++)
            {
                int cost;
                cost = abs(man[i].x-house[j].x)+abs(man[i].y-house[j].y);
                addedge(i+1,j+msize+1,1,0,cost);
            }
        }
        for(j = 0;j < man.size();j++)
        {
            addedge(0,j+1,1,0,0);
        }
        for(i = 0;i < hsize;i++)
        addedge(msize+1+i,msize+hsize+1,1,0,0);

        int ans;
        ans = 0;
        ans = mcmf(0,hsize+msize+1,hsize+msize+2);
        cout<<ans<<endl;
    }
    return 0;
}