POJ 2195 Going Home(费用流)

POJ 2195 Going Home(费用流)

http://poj.org/problem?id=2195

题意:

   给定一个N*M的地图，地图上有若干个man和house，且man与house的数量一致。man每移动一格需花费$1（即单位费用=单位距离），一间house只能入住一个man。现在要求所有的man都入住house，求最小费用。

分析:

   之前用二分图最大权匹配做过一次这个题目:

http://blog.csdn.net/u013480600/article/details/38735423

   现在用费用流再做一次,建图如下:

   源点s编号0, 人编号1到n, 房子编号n+1到n+n, 汇点编号t.

   源点s到每个人i有边(s, i, 1,0)

   每个人i到每个房子j有边(i, j, 1, i人到j房的开销)

   每个房子j到汇点t有边(j, t, 1, 0)

   最终我们求出的最小费用就是所求.

AC代码:

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#include<vector>#include<cmath>#define INF 1e9using namespace std;const int maxn=200+5;struct Edge{    int from,to,cap,flow,cost;    Edge(){}    Edge(int f,int t,int c,int fl,int co):from(f),to(t),cap(c),flow(fl),cost(co){}};struct MCMF{    int n,m,s,t;    vector<Edge> edges;    vector<int> G[maxn];    bool inq[maxn];    int d[maxn];    int p[maxn];    int a[maxn];    void init(int n,int s,int t)    {        this->n=n, this->s=s, this->t=t;        edges.clear();        for(int i=0;i<n;++i) G[i].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 &flow,int &cost)    {        for(int i=0;i<n;++i) d[i]=INF;        queue<int> Q;        memset(inq,0,sizeof(inq));        d[s]=0, Q.push(s), a[s]=INF, p[s]=0, inq[s]=true;        while(!Q.empty())        {            int u=Q.front(); Q.pop();            inq[u]=false;            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]=true; }                }            }        }        if(d[t]==INF) return false;        flow +=a[t];        cost +=a[t]*d[t];        int u=t;        while(u!=s)        {            edges[p[u]].flow +=a[t];            edges[p[u]^1].flow -=a[t];            u=edges[p[u]].from;        }        return true;    }    int solve()    {        int flow=0,cost=0;        while(BellmanFord(flow,cost));        return cost;    }}MM;struct Node{    int x,y;    Node(){}    Node(int x,int y):x(x),y(y){}    int get_dist(Node& b)    {        return abs(x-b.x)+abs(y-b.y);    }}node1[maxn],node2[maxn];int main(){    int n,m;    while(scanf("%d%d",&n,&m)==2 && n)    {        int num1=0,num2=0;//记录人数和房子数        for(int i=1;i<=n;++i)        for(int j=1;j<=m;++j)        {            char ch;            scanf(" %c",&ch);            if(ch=='m') node1[num1++]=Node(i,j);            else if(ch=='H') node2[num2++]=Node(i,j);        }        int src=0,dst=num1*2+1;        MM.init(num1*2+2,src,dst);        for(int i=1;i<=num1;++i)        {            MM.AddEdge(src,i,1,0);            MM.AddEdge(num1+i,dst,1,0);            for(int j=1;j<=num1;++j)            {                MM.AddEdge(i,num1+j,1,node1[i-1].get_dist(node2[j-1]));            }        }        printf("%d\n",MM.solve());    }    return 0;}