POJ - 2195 Going Home （最小费用最大流）

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

题解：

把每个人到每个房子的距离当成费用，添加超级源点超级汇点，所有边的流量全部为1建立模型。然后就是基础的最小费用最大流。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>

using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 310; 

struct Edge{
	int value,flow,to,rev;
	Edge(){}
	Edge(int a,int b,int c,int d):to(a),value(b),flow(c),rev(d){}
};

vector<Edge> E[MAXN];

inline void Add(int from,int to,int flow,int value){
	E[from].push_back(Edge(to,value,flow,E[to].size()));
	E[to].push_back(Edge(from,-value,0,E[from].size()-1));
}

struct House{
	int x,y;//坐标 
	House(int a,int b):x(a),y(b){}
};

vector<House> H;

struct People{
	int x,y;
	People(int a,int b):x(a),y(b){}
};

vector<People> P;

bool book[MAXN];//用于SPFA中标记是否在queue中 
int cost[MAXN];//存费用的最短路径 
int pre[MAXN];//存前节点 
int pree[MAXN];//存在前节点的vector中的下标 

bool Spfa(int from,int to){
	memset(book,false,sizeof book);
	memset(cost,INF,sizeof cost);
	book[from] = true;
	cost[from] = 0;
	queue<int> Q;
	Q.push(from);
	while(!Q.empty()){
		int t = Q.front();
		book[t] = false;
		Q.pop();
		for(int i=0 ; i<E[t].size() ; ++i){
			Edge& e = E[t][i];
			if(e.flow > 0 && cost[e.to] > cost[t] + e.value){
				cost[e.to] = cost[t] + e.value;
				pre[e.to] = t;
				pree[e.to] = i;
				if(book[e.to] == false){
					Q.push(e.to);
					book[e.to] = true;
				}
			}
		}
	}
	return cost[to] != INF;
}



int Work(int from,int to){
	int sum = 0;
	while(Spfa(from,to)){
		int mflow = INF;//SPFA找到的最短路径的最小容量 
		int flag = to;
		while(flag != from){
			mflow = min(mflow,E[pre[flag]][pree[flag]].flow);
			flag = pre[flag];
		}
		flag = to;
		while(flag != from){
			sum += E[pre[flag]][pree[flag]].value * mflow;
			E[pre[flag]][pree[flag]].flow -= mflow;
			E[flag][E[pre[flag]][pree[flag]].rev].flow += mflow;
			flag = pre[flag];
		}
	}
	return sum;
}

int main(){
	
	int N,M;
	char ch;
	while(scanf("%d %d",&N,&M) && (N || M)){
		for(int i=1 ; i<=N ; ++i){
			getchar();
			for(int j=1 ; j<=M ; ++j){
				ch = getchar();
				if(ch == 'H')H.push_back(House(i,j));
				else if(ch == 'm')P.push_back(People(i,j));
			}
		}
		for(int i=0 ; i<P.size() ; ++i)Add(0,i+1,1,0);
		for(int i=0 ; i<H.size() ; ++i)Add(i+100+1,MAXN-1,1,0);
		for(int i=0 ; i<P.size() ; ++i){
			for(int j=0 ; j<H.size() ; ++j){
				int value = abs(P[i].x - H[j].x) + abs(P[i].y - H[j].y);
				Add(i+1,j+100+1,1,value);
			}
		}
		printf("%d\n",Work(0,MAXN-1));//超级源点和超级汇点为0和MAXN-1 
		for(int i=0 ; i<MAXN ; ++i)E[i].clear();
		P.clear();
		H.clear();
	}
	
	return 0;
}