zoukankan      html  css  js  c++  java
  • Poj 2195 Going Home(费用流)

    Going Home
    Time Limit: 1000MS Memory Limit: 65536K
    Total Submissions: 22327 Accepted: 11280
    Description
    这里写图片描述
    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.
    Input
    There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of ‘H’s and ‘m’s on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.
    Output
    For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.
    Sample Input
    2 2
    .m
    H.
    5 5
    HH..m
    …..
    …..
    …..
    mm..H
    7 8
    …H….
    …H….
    …H….
    mmmHmmmm
    …H….
    …H….
    …H….
    0 0
    Sample Output
    2
    10
    28
    Source
    Pacific Northwest 2004

    /*
    费用流裸题.
    二分图搞.
    从源点向man连边流量为1费用为0.
    从man向house连边流量为1费用为曼哈顿距离.
    从house向汇点连边流量为1费用为0. 
    */
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<queue>
    #define MAXN 101
    #define INF 1e9
    using namespace std;
    int n,m,tot1,tot2,cut=1,S,T,ans;
    int g[MAXN][MAXN],vis[MAXN*MAXN],fa[MAXN*MAXN],head[MAXN*MAXN],dis[MAXN*MAXN];
    struct data{int x,y;}a[MAXN*MAXN],b[MAXN*MAXN];
    struct edge{int u,v,c,f,next;}e[MAXN*MAXN*2];
    queue<int>q;
    void add(int u,int v,int c,int f)
    {
        e[++cut].u=u;e[cut].v=v;e[cut].next=head[u];e[cut].c=c;e[cut].f=f;head[u]=cut;
        e[++cut].u=v;e[cut].v=u;e[cut].next=head[v];e[cut].c=0;e[cut].f=-f;head[v]=cut;
    }
    bool bfs(int t)
    {
        for(int i=1;i<=T;i++) dis[i]=INF;dis[0]=0;
        q.push(S);
        while(!q.empty())
        {
            int u=q.front();q.pop();vis[u]=0;
            for(int i=head[u];i;i=e[i].next)
            {
                int v=e[i].v;
                if(dis[v]>dis[u]+e[i].f&&e[i].c)
                {
                    dis[v]=dis[u]+e[i].f;fa[v]=i;
                    if(vis[v]!=t) vis[v]=t,q.push(v);
                }
            }
        }
        return dis[T]!=INF;
    }
    void mincost()
    {
        int t=1;ans=0;
        while(bfs(t))
        {
            int tmp=fa[T],x=INF;
            while(tmp) x=min(x,e[tmp].c),tmp=fa[e[tmp].u];
            tmp=fa[T];
            while(tmp)
            {
                e[tmp].c-=x;
                e[tmp^1].c+=x;
                e[tmp].f*=x;
                ans+=e[tmp].f;
                tmp=fa[e[tmp].u];
            }
            t++;
        }
        printf("%d
    ",ans);
    }
    void slove()
    {
        for(int i=1;i<=tot1;i++) add(0,i,1,0);
        for(int i=1;i<=tot2;i++) add(i+tot1,tot1+tot2+1,1,0);
        for(int i=1;i<=tot1;i++)
          for(int j=1;j<=tot2;j++)
          {
            int x=fabs(a[i].x-b[j].x)+fabs(a[i].y-b[j].y);
            add(i,tot1+j,1,x);
          }
        mincost();
        return ;
    }
    void Clear()
    {
        tot1=tot2=0;cut=1;
        memset(head,0,sizeof head);
        memset(b,0,sizeof b);
        return ;
    }
    int main()
    {
        char ch;
        while(scanf("%d%d",&n,&m))
        {
            if(!n&&!m) break;
            Clear();
            for(int i=1;i<=n;i++)
              for(int j=1;j<=m;j++)
              {
                cin>>ch;
                if(ch=='m') a[++tot1].x=i,a[tot1].y=j;
                else if(ch=='H') b[++tot2].x=i,b[tot2].y=j;
              }
            T=tot1+tot2+1;slove();
        }
        return 0;
    }
  • 相关阅读:
    远程连接mysql数据库注意点记录
    一个支持chrome、firefox的全屏插件
    Windows Phone应用程序Tombstone执行模型总结
    Windows Phone页面导航和独立存储开发总结
    【讨论帖】控制分布式缓存“及时”过期的一种实现
    Java HashMap的死循环的启示
    You must use the Role Management Tool to install or configure Microsoft .NET Framework 3.5 SP1
    Analysis Service Tabular Model #003 Multidimensional Model VS Tabular Model 我们该如何选择?
    Analysis Service Tabular Model #002 Analysis services 的结构:一种产品 两个模型
    基于DotNet构件技术的企业级敏捷软件开发平台 AgileEAS.NET 4.0 最新发布版本 下载使用说明
  • 原文地址:https://www.cnblogs.com/nancheng58/p/10068077.html
Copyright © 2011-2022 走看看