[poj] 2195 Going Home || 最小费用最大流

Posted 2020-10-16 Mrha

原题

给定一个N*M的地图，地图上有若干个人和房子，且人与房子的数量一致。人每移动一格需花费1（即单位费用=单位距离），一间房子只能入住一个人。现在要求所有的人都入住，求最小费用。

把每个人和每个房子连费用为距离，容量为1的边就可以了。

#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#define N 220
#define inf 0x3f3f3f3f
using namespace std;
int n,m,head[N],dis[N],cur[N],ans,cnt=2,s,t,ANS,T,cse,xm[110],ym[110],xh[110],yh[110],man,house;
queue <int> q;
char a;
bool vis[N];
struct hhh
{
    int to,next,w,cost;
}edge[50005];

int read()
{
    int ans=0,fu=1;
    char j=getchar();
    for (;(j<'0' || j>'9') && j!='-';j=getchar()) ;
    if (j=='-') fu=-1,j=getchar();
    for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0';
    return ans*fu;
}

void add(int u,int v,int w,int c)
{
    edge[cnt].to=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    edge[cnt].cost=c;
    head[u]=cnt++;
}

void addEdge(int u,int v,int w,int c)
{
    add(u,v,w,c);
    add(v,u,0,-c);
}

bool bfs()
{
    for (int i=s;i<=t;i++)
    vis[i]=0,cur[i]=head[i],dis[i]=inf;
    q.push(s);
    dis[s]=0;
    vis[s]=1;
    while(!q.empty())
    {
    int r=q.front();
    q.pop();
    vis[r]=0;
    for (int i=head[r],v;i;i=edge[i].next)
    {
        v=edge[i].to;
        if (edge[i].w>0 && dis[r]+edge[i].cost<dis[v])
        {
        dis[v]=dis[r]+edge[i].cost;
        if (!vis[v])
        {
            vis[v]=1;
            q.push(v);
        }
        }
    }
    }
    return dis[t]!=inf;
}

int dfs(int x,int f)
{
    if (x==t) return ANS+=f*dis[t],f;
    int ha=0,now;
    vis[x]=1;
    for (int &i=cur[x],v;i;i=edge[i].next)
    {
    v=edge[i].to;
    if (vis[v]) continue;
    if (edge[i].w>0 && dis[v]==dis[x]+edge[i].cost)
    {
        now=dfs(v,min(f-ha,edge[i].w));
        if (now)
        {
        ha+=now;
        edge[i].w-=now;
        edge[i^1].w+=now;
        }
    }
    if (ha==f) return ha;
    }
    return ha;
}

void init()
{
    memset(head,0,sizeof(head));
    cnt=2;
    ANS=0;
    ans=0;
    man=0;
    house=0;
}

int main()
{
    while (~scanf("%d%d",&n,&m))
    {
    if (!n && !m) break;
    init();
    s=0;
    for (int i=1;i<=n;i++)
    {
        getchar();
        for (int j=1;j<=m;j++)
        {
        a=getchar();
        if (a=='m') xm[++man]=i,ym[man]=j;
        else if (a=='H') xh[++house]=i,yh[house]=j;
        }
    }
    t=man+house+1;
    for (int i=1;i<=man;i++)
        addEdge(s,i,1,0);
    for (int i=1;i<=house;i++)
        addEdge(man+i,t,1,0);
    for (int i=1;i<=man;i++)
        for (int j=1;j<=house;j++)
        addEdge(i,man+j,1,abs(xh[j]-xm[i])+abs(yh[j]-ym[i]));
    while (bfs()) ans+=dfs(s,inf);
    printf("%d\n",ANS);
    }
    return 0;
}