POJ Going Home (2195)(最小费用最大流)

Going Home

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 22076 Accepted: 11148

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.mH.5 5HH..m...............mm..H7 8...H.......H.......H....mmmHmmmm...H.......H.......H....0 0

Sample Output

21028

Source

Pacific Northwest 2004

题意：有n*m的矩阵，H表示这个点是一个房子，m表示这个点是一个人，现在每一个人需要走入一个房间，已经知道的是认得数目和房子的个数一定是相同的，现在问这些人都回到一个房间所走的总的步数最小

分析：找这些人到这些房子的最小步数，我们可以建一个超级源点和超级汇点，源点到每个人的费用为0，距离为1

每个房子到汇点费用为0，距离为1，人到房子的距离是1，费用为坐标差的和

//数组开大一些

#include <stdio.h>#include <string.h>#include <queue>#include <stdlib.h>using namespace std;#define inf 0x3f3f3f3f#define N 300struct node{    int u,v,cap,cost,next;}eg[N*N];struct node1{    int x,y;}q1[N*N],q2[N*N];char mp[200][200];int head[N],vis[N],dis[N],pre[N];int top;void add(int u,int v,int cap,int cost){    eg[top].u=u;    eg[top].v=v;    eg[top].cap=cap;    eg[top].cost=cost;    eg[top].next=head[u];    head[u]=top++;    eg[top].u=v;    eg[top].v=u;    eg[top].cap=0;    eg[top].cost=-cost;    eg[top].next=head[v];    head[v]=top++;}int spfa(int s,int t){    int i,j;    for(i=1;i<=t;i++)    {        dis[i]=inf;        pre[i]=-1;        vis[i]=0;    }    int q[112345];    int in=0,out=0;    q[in++]=s;    dis[s]=0;    vis[s]=1;    while(in>out)    {        int u=q[out++];        vis[u]=0;        for(i=head[u];i!=-1;i=eg[i].next)        {            if(eg[i].cap>0)            {                int v=eg[i].v;                if(dis[v]>dis[u]+eg[i].cost)                {                    dis[v]=dis[u]+eg[i].cost;                    pre[v]=i;   //i指的是以v为终点这条边在eg中的编号                    if(!vis[v])                    {                        vis[v]=1;                        q[in++]=v;                    }                }            }        }    }    if(dis[t]==inf)return 0;    else return dis[t];}int sap(int s,int t){    int i;    int ans=0,max_flow=0;    int flow;    while(spfa(s,t))    {        for(i=1;i<=t;i++)            printf("%d  %d \n",i,pre[i]);        top=0;        flow=inf;        for(i=pre[t];i!=-1;i=pre[eg[i].u])            if(eg[i].cap<flow)            flow=eg[i].cap;        for(i=pre[t];i!=-1;i=pre[eg[i].u])        {            eg[i].cap-=flow;            eg[i^1].cap+=flow;        }        ans+=dis[t];        max_flow+=flow;    }    return ans;}int main(){    int n,m,i,j;    while(~scanf("%d%d",&n,&m))    {        top=0;        memset(head,-1,sizeof(head));        int top1=0,top2=0;        if(n==0&&m==0)            break;        for(i=0;i<n;i++)        {            scanf("%s",mp[i]);        }        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                if(mp[i][j]=='m')                {                    q1[top1].x=i;                    q1[top1++].y=j;                }                else if(mp[i][j]=='H')                {                    q2[top2].x=i;                    q2[top2++].y=j;                }            }        }        for(i=0;i<top1;i++)        {            add(1,i+2,1,0);            for(j=0;j<top2;j++)            {                int l=abs(q1[i].x-q2[j].x)+abs(q1[i].y-q2[j].y);                //printf("   %d\n",l);                add(i+2,top1+j+2,1,l);            }        }        for(j=0;j<top2;j++)            add(top1+j+2,top1+top2+2,1,0);        int s=1,t=top1+top2+2;        printf("%d\n",sap(s,t));    }    return 0;}