二分图完备匹配(最小费用 || KM) poj 2195 GoingHome

Going Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9246		Accepted: 4765

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

2





10





28

最小花费最大流解法：



/*
题目大意：n个人到m个房子分布在在坐标轴的不同位置,求解n个人到m间房子的最小花费(人可以上下左右移动,每移动一次就花费$1)；
题目解答：这是一道二分最大匹配问题，由于题目特殊性(n==m)最大匹配就是完备匹配.接下来就是转化成二分图的问题,
		：将n个人归为X集合m个房屋归为Y集合,增加一个源点和汇点,寻找一条最小花费的最大匹配
		：（可以采用队列实现的bellman算法(SPFA)寻找一条最短路的增广路径）代替dfs寻找的增广路径就ok了.
代码技巧：struct {i,j}记录坐标,便于后面计算距离(花费).dis[mi][mi+hi]就是第i个人到第hi个房子的距离(花费),
		：c[x][y]=0,1表示参与流量也因为都只有0和1.d[x]当前增广路径中源点S到x点的当前最短距离.
*/
Source Code

Problem: 2195   User: wawadimu 
Memory: 608K   Time: 79MS 
Language: C++   Result: Accepted 


Source Code 
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

#define inf INT_MAX
#define maxn 110
struct node
{
	int i,j;//记录坐标
}man[maxn],house[maxn];

//dis[x][y]顶点x到顶点y的距离
//c[x][y]表示当前残余流量值:1或者0
//p[x]记录增益路径中x点前一个顶点
//inq[x]标记函数
//hi房子数目,mi人数
//增益路径中d[x]是S到x的最短距离
int dis[2*maxn][2*maxn],c[2*maxn][2*maxn];
int p[2*maxn];
bool inq[2*maxn];
int d[2*maxn];
int hi,mi;//mi==hi
//采用FIFO队列Bellman_Ford寻找一条增益路径,成功返回1，失败返回0
int bellman_ford(int s,int t)
{
	queue<int> q;
	q.push(s);
	//初始化标记函数和距离函数
	memset(inq,0,sizeof(inq));
	for(int i=0;i<=2*mi+1;i++)  d[i]=inf;
	d[s]=0;
	inq[s]=1;
	while(!q.empty())
	{
		int x=q.front();
            q.pop();
		inq[x]=false;
		for(int v=1;v<=2*mi+1;v++)
		{
			if(d[v] > d[x] + dis[x][v] && c[x][v]) //有残余,松弛
			{
				d[v]=d[x]+dis[x][v];
				p[v]=x;                            //记录路径
				if(!inq[v])                        //已经在队列中就不需要重复添加
				{
					inq[v]=true;
					q.push(v);
				}
			}
		}
	}
	return d[2*mi+1]!=inf;//是否找到增益路径
}

int main()
{
	//freopen("2195.txt","r",stdin);
	//freopen("out1.txt","w",stdout);
	int n,m;
	int i,j;
	char ch;
	//int mi,hi;
	while(scanf("%d%d/n",&n,&m)!=EOF && !(n==0 && m==0))
	{
		mi=hi=1;
		//scanf("&c",&ch);
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				//cin>>ch;
				scanf("%c/n",&ch);
				if(ch=='m')
				{
					man[mi].i = i;
					man[mi++].j = j;
				}
				if(ch=='H')
				{
					house[hi].i = i;
					house[hi++].j = j;
				}

			}
		}
		hi--;mi--;
		//初始化残余流量为0
		//dis[x][y] x,y不相连
		memset(c,0,sizeof(c));
		for(i=0;i<=2*mi+1;i++)
			for(j=0;j<=2*mi+1;j++)
				dis[i][j]=inf;
		for(i=1;i<=mi;i++)
		{
			for(j=1;j<=hi;j++)
			{
				//计算dis[man][house]=|i|+|j|;
				//计算c[man][house]=1;
				dis[i][mi+j]=abs(man[i].i-house[j].i) + abs(man[i].j - house[j].j);
				dis[mi+j][i]=-dis[i][mi+j];
				c[i][mi+j]=1;
			}
		}
		for(i=1;i<=mi;i++)
		{
			//创造一个源点S[0]和汇点t[2*mi+1]
			//dis[0][man]=dis[man][0]=0 c[0][man]=1;
			//dis[t][house]=dis[house][t]=0 c[house][t]=1;
			dis[0][i]=dis[i][0]=0;
			dis[mi+i][2*mi+1]=dis[2*mi+1][mi+i]=0;
			c[0][i]=1;
			c[mi+i][2*mi+1]=1;
		}
		int s=0,t=2*mi+1;
		int min=0;//记录费用
		while(bellman_ford(s,t))
		{
			for(int u=t;u!=s;u=p[u])
			{
				//cout<<u<<" ";
				c[p[u]][u]-=1;
				c[u][p[u]]+=1;
			}
			min+=d[2*mi+1];
		}
		cout<<min<<endl;
	}
	return 1;
}






















KM解法：





/*
题目大意：略；
题目解答：KM算法。前提条件：加权完全二分图(两个子集的任意两个顶点都相关联)。相等子图满足lx(i)+ly(j)=w(i,j);
		：(1)扩展相等子图G(l),在G(l)中寻找一条增广路径
		：首先将出发顶点s加入到想等子图G(l)中，然后依次搜索顶点s出发的每一条边(s,j)，如果s和j的顶标和等于权值
		：则j进去G(l)，若j是未盖点或存在j出发的增广路径，则(s,j)是匹配边，退出
		：(2)调整下标
		：搜索所有X集合中每个满足下列条件：i属于G(l),j不属于G(l)；
		：min=minmum(map[i][j]-lx[i]+lx[j]); lx[i]+=min && ly[j属于G(l)]-=min;(对于已经匹配的顶点lx[x]+ly[y]==map[x][y],
		：也就是不影响原先匹配的相等子图)
		：在寻找增广路径时，KM算法不断的调整顶标，直到找到一条增广路径。由于匹配数目M的不断增加，增广过程的信息都保存。
		：每一次增广过程最坏复杂度o(E),总的复杂度O(NE^2);

 
Source Code

Problem: 2195  User: wawadimu 
Memory: 224K  Time: 0MS 
Language: C++  Result: Accepted 

Source Code 
*/
#include<iostream>
using namespace std;

#define inf INT_MAX
#define maxn 110
struct point
{
	int x,y;
}man[maxn],house[maxn];
int N,M;
int m,h;
int map[maxn][maxn];
int match[maxn];//match[y]=x Y集合匹配的X集合下标
int vx[maxn],vy[maxn];//标记数组
int lx[maxn],ly[maxn];//定标

bool dfs(int i)//深度优先遍历寻找增益路径
{
	vx[i]=1;
	for(int j=1;j<=h;j++)
	{
		if(!vy[j] && lx[i]+ly[j]== map[i][j])
		{
			vy[j]=1;
			if(match[j]==-1 || dfs(match[j]))
			{
				match[j]=i;
				return true;
			}
		}
	}
	return false;
}
int main()
{
	//freopen("2195.txt","r",stdin);
	int i,j,k;
	char ch;
	while(scanf("%d%d/n",&N,&M)!=EOF && !(N==0 && M==0))
	{
		m=h=0;
		for(i=1;i<=N;i++)
		{
			for(j=1;j<=M;j++)
			{
				scanf("%c",&ch);
				//cout<<ch;
				if(ch=='m')
				{
					man[++m].x=i;
					man[m].y=j;
				}
				else if(ch=='H')
				{
					house[++h].x=i;
					house[h].y=j;
				}
				scanf("/n");//过滤换行符
			}
			//cout<<endl;
		}
		//memset(map,inf,sizeof(map));
		memset(match,-1,sizeof(match));
		for(i=1;i<=m;i++)
		{
			for(j=1;j<=h;j++)
			{
				map[i][j]=abs(man[i].x-house[j].x)+abs(man[i].y- house[j].y);
			}
		}
		for(i=1;i<=m;i++)
		{
			lx[i]=inf;
			for(j=1;j<=h;j++)
			{
				if(lx[i] > map[i][j]) 
					lx[i]=map[i][j];
			}
			//cout<<lx[i]<<endl;
		}//最小权匹配
		memset(ly,0,sizeof(ly));
		int min;
		int cnt=0;
		for(i=1;i<=m;i++)//注意下标变量
		{
			while(1)
			{
				memset(vx,0,sizeof(vx));
				memset(vy,0,sizeof(vy));
				if(dfs(i)) break;
				min=inf;
				for(j=1;j<=m;j++)
				{
					if(vx[j])
					{
						for(k=1;k<=h;k++)
						{
							if(!vy[k] && map[j][k] - lx[j]-ly[k] <min)
							{
								min=map[j][k] - lx[j]-ly[k];
							}
						}
					}
				}
				for(j=1;j<=m;j++) if(vx[j]) lx[j]+=min;
				for(j=1;j<=h;j++) if(vy[j]) ly[j]-=min;
			}
		}
		int ans=0;
		for(i=1;i<=h;i++)
		{
			ans+=map[match[i]][i];
		}
		printf("%d/n",ans);
	}
	return 0;
}