寒假Day14：最小费用最大流问题 POJ2195-Going Home

POJ2195-Going Home

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.

InputThere 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.
OutputFor 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

 题意：

　　给出n、m，n行m列，‘ . ’代表空地，H代表房子，m代表man，有同样数目的人和房子（数量≤100），每个人对应一座房子；

　　问：最少需要花费多少美元（每个人需要走的步数）才能使得每个人都到房子里去（一人只能进入一所）；

　　间接性问：最小费用最大流。

思路：

• 源点s到每个人建立一条容量为1,费用为0的边；
• 每个人到每个房子建立一条容量为1,费用为两者的距离；
• 每个房子到汇点t建立一条容量为1,费用为0的边；
• 计算走的步数：曼哈顿距离
• 也可以用二分图匹配（KM算法来做）

 

 

 

注意：

1. 数组开的大小；
2. 学会如何建边；
3. 看清字母啊大哥，是H和m，看清大小写啊你个猪

 

难点还是在于建边：

s=0; //源点s
t=hh+mm+1; //汇点

for(int i=1; i<mm; i++)
    add(s,i,1,0);//源点s到每个人建立一条容量为1,费用为0的边

for(int i=1; i<mm; i++)
{
    for(int j=1; j<hh; j++)
    {
        int dd=d(H[i].x,H[i].y,M[j].x,M[j].y);
        add(i,j+hh,1,dd);//每个人到每个房子建立一条容量为1,费用为两者的距离边
    }
}

for(int i=1; i<hh; i++)
    add(i+hh,t,1,0);//每个房子到汇点t建立一条容量为1,费用为0的边

 

AC代码：

#include<string.h>
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
const int N=1100;
typedef long long ll;

struct node
{
    int to,nextt;
    int cap,flow,cost;
    int x,y;
} e[N*N],H[N*N],M[N*N];

bool book[N*N];
int pre[N*N],head[N*N],dist[N*N];
int tot,s,t;
char a[N][N];

int d(int x1,int y1,int x2,int y2)
{
    return abs(x2-x1)+abs(y2-y1);
}

void add(int u,int v,int cap,int cost)
{
    e[tot].to=v;
    e[tot].cap=cap;
    e[tot].cost=cost;
    e[tot].flow=0;
    e[tot].nextt=head[u];

    head[u]=tot++;
    e[tot].to=u;
    e[tot].cap=0;
    e[tot].cost=-cost;
    e[tot].flow=0;
    e[tot].nextt=head[v];
    head[v]=tot++;
}

bool SPFA()
{
    for(int i=0; i<=t; i++)
    {
        dist[i]=inf;
        book[i]=0;
        pre[i]=-1;
    }
    book[s]=1;
    dist[s]=0;
    queue<int>Q;
    Q.push(s);
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        book[u]=0;
        for(int i=head[u]; i!=-1; i=e[i].nextt)
        {
            int v=e[i].to;
            if(e[i].cap>e[i].flow&&dist[v]>dist[u]+e[i].cost)
            {
                dist[v]=dist[u]+e[i].cost;
                pre[v]=i;
                if(book[v]==0)
                {
                    book[v]=1;
                    Q.push(v);
                }
            }
        }
    }
    if(dist[t]!=inf)
        return 1;
    return 0;
}

int MCMF()
{
    int flow=0,cost=0;
    while(SPFA())
    {
        int minn=inf;
        for(int i=pre[t]; i!=-1; i=pre[e[i^1].to])
            minn=min(minn,e[i].cap-e[i].flow);
        for(int i=pre[t]; i!=-1; i=pre[e[i^1].to])
        {
            e[i].flow+=minn;
            e[i^1].flow-=minn;
            cost+=e[i].cost*minn;
        }
        flow+=minn;
    }
    return cost;
}

int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        memset(head,-1,sizeof(head));
        int hh=1,mm=1;
        for(int i=0; i<n; i++)
        {
            scanf("%s",a[i]);
            for(int j=0;j<m;j++)
            {
                if(a[i][j]=='H')//home
                {
                    H[hh].x=i;
                    H[hh++].y=j;

                }
                else if(a[i][j]=='m')//man
                {
                    M[mm].x=i;
                    M[mm++].y=j;
                }
            }
        }

        s=0;
        t=hh+mm+1;
        for(int i=1;i<mm;i++)
            add(s,i,1,0);//源点s到每个人建立一条容量为1,费用为0的边

        for(int i=1;i<mm;i++)
        {
            for(int j=1;j<hh;j++)
            {
                int dd=d(H[i].x,H[i].y,M[j].x,M[j].y);
                add(i,j+hh,1,dd);//每个人到每个房子建立一条容量为1,费用为两者距离的边
            }
        }

        for(int i=1;i<hh;i++)
            add(i+hh,t,1,0);//每个房子到汇点t建立一条容量为1,费用为0的边

        int mincost=MCMF();
        printf("%d\n",mincost);
    }
    return 0;
}

 

 

---

 

 

待解决：

• POJ2195数组大小问题：

这题的数组到底应该开多大啊？？？好不容易试过一组，可是不知道数组怎么确定大小啊？？？总不能每次都试叭叭叭

我觉得应该是这样子的，我的理由是从源点s到所有的人连N条边，人和房子连N*N条边，房子到汇点t连N条边，所有边加起来=12320，可是这样不对，

邻接表开4倍、10倍、100倍都RT了，

为什么？？？

//12320=N+N*N+N=110+110*110+110
const int N=12310;
struct node
{
    int to,nextt;
    int cap,flow,cost;
    int x,y;
} e[110*110],H[110],M[110];

bool book[N];
int pre[10*N],head[10*N],dist[N];
int tot,s,t;
char a[110][110];

 

我是这样写才试过的（已AC）

const int N=1100;

struct node
{
    int to,nextt;
    int cap,flow,cost;
    int x,y;
} e[N*N],H[N*N],M[N*N];

bool book[N*N];
int pre[N*N],head[N*N],dist[N*N];
int tot,s,t;
char a[N][N];