HDU 1533 Going Home 最小费用最大流

最新推荐文章于 2019-02-18 17:32:59 发布

原创最新推荐文章于 2019-02-18 17:32:59 发布 · 444 阅读

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#网络流 #最小费用最大流

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本文介绍了一个经典的最小费用最大流问题实例，通过构建网络图来解决小人与房屋匹配的问题，旨在寻找从所有小人到不同房屋的路径中总花费最小的方案。文章详细解释了建图的方法及具体的实现代码。

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Description

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

21028

一看就是网络流水题，建图很简单，源点到每个人建一条流量为1的边，费用为0，每个人到每个house建一条边，流量为1，费用为两点之间的距离，然后每个house到汇点建一条流量为1费用为0的边即可。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define maxn 500
#define inf 0x3f3f3f
using namespace std;
struct node
{
    int st;
    int en;
    int flow,cost;
    int next;
}E[101000];
int num;
int p[maxn];
void init()
{
    memset(p,-1,sizeof p);
    num=0;
}
void add(int st,int en,int flow,int cost)
{
    E[num].st=st;
    E[num].en=en;
    E[num].flow=flow;
    E[num].cost=cost;
    E[num].next=p[st];
    p[st]=num++;
    E[num].st=en;
    E[num].en=st;
    E[num].flow=0;
    E[num].cost=-cost;
    E[num].next=p[en];
    p[en]=num++;
}
int pre[maxn];
int dis[maxn];
bool fg[maxn];
bool spfa(int st,int en)
{
    for(int i=0;i<=en;i++)
        fg[i]=0,dis[i]=inf,pre[i]=-1;
    queue<int>q;
    q.push(st);
    fg[st]=1;
    dis[st]=0;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        fg[u]=0;
        for(int i=p[u];i+1;i=E[i].next)
        {
            int v=E[i].en;
            if(E[i].flow&&dis[v]>dis[u]+E[i].cost)
            {
                dis[v]=dis[u]+E[i].cost;
                pre[v]=i;
                if(!fg[v])
                {
                    fg[v]=1;
                    q.push(v);
                }
            }
        }
    }
    if(dis[en]<inf)
        return 1;
    return 0;
}
int solve(int st,int en)
{
    int ans=0;
    while(spfa(st,en))
    {
        int d=inf;
        for(int i=pre[en];i+1;i=pre[E[i].st])
            d=min(d,E[i].flow);
        for(int i=pre[en];i+1;i=pre[E[i].st])
        {
            E[i].flow-=d;
            E[i^1].flow+=d;
            ans+=d*E[i].cost;
        }
    }
    return ans;
}
char ma[102][102];
struct Node
{
    int x,y;
}mnum[102],hnum[102];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m),n||m)
    {
        init();
        int s=0,t=300;
        for(int i=0;i<n;i++)
            scanf("%s",ma[i]);
        int km=0,kh=0;///统计m和h的个数
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
            {
                if(ma[i][j]=='m')
                {
                    km++;
                    mnum[km].x=i;
                    mnum[km].y=j;
                }
                else if(ma[i][j]=='H')
                {
                    kh++;
                    hnum[kh].x=i;
                    hnum[kh].y=j;
                }
            }
        for(int i=1;i<=km;i++)
        {
            add(s,i,1,0);
            for(int j=1;j<=kh;j++)
            {
                add(i,j+km,1,abs(mnum[i].x-hnum[j].x)+abs(mnum[i].y-hnum[j].y));
            }
        }
        for(int i=1;i<=kh;i++) add(i+km,t,1,0);
        printf("%d\n",solve(s,t));
    }
    return 0;
}