POJ 2195 Going Home (最小費用最大流)

_TCgogogo_發表於2016-03-25
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Going Home
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 20617   Accepted: 10449

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

Source

Pacific Northwest 2004

題目連結：http://poj.org/problem?id=2195

題目大意：n個人n個房子，每間房只能有一個人，每個人走一步花費1元，求每個人都到房子裡所花費的最小總費用

題目分析：基礎的最小費用流，建圖(除了超源和超匯其實是個完全二分圖)：超級源點到每個man容量為1，費用為0，每個man到每個house容量為1，費用為兩點間的曼哈頓距離，每個house到超級匯點容量為1，費用為0，跑一下最小費用流即可
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cstdlib>
#define x first
#define y second
#define mk make_pair
using namespace std;
int const INF = 0x3fffffff;
int const MAX = 205;
int n, m;
char a[105][105];
int head[MAX], cnt;
int pre[MAX], vis[MAX], dis[MAX];
int src, sk, ans;

struct EGDE
{
	int to, nxt, cap, cost;
}e[MAX * MAX / 2];

pair < int, int > house[105];
pair < int, int > man[105];
int hnum, mnum;

void Init()
{
	hnum = 1;
	mnum = 1; 
	cnt = 0;
	ans = 0;
	memset(head, -1, sizeof(head));
}

void Add(int u, int v, int cap, int cost)
{
	e[cnt].to = v;
	e[cnt].cap = cap;
	e[cnt].cost = cost;
	e[cnt].nxt = head[u];
	head[u] = cnt ++;

	e[cnt].to = u;
	e[cnt].cap = 0;
	e[cnt].cost = -cost;
	e[cnt].nxt = head[v];
	head[v] = cnt ++;
}

int Get_dis(int i, int j)
{
	return abs(house[i].x - man[j].x) + abs(house[i].y - man[j].y);
}

void Build_graph()
{
	for(int i = 1; i <= n; i++)
	{
		scanf("%s", a[i] + 1);
		for(int j = 1; j <= m; j++)
		{
			if(a[i][j] == 'H')
				house[hnum ++] = mk(i, j);
			if(a[i][j] == 'm')
				man[mnum ++] = mk(i, j);
		}
	}
	src = 0;
	sk = hnum + mnum - 1;
	for(int i = 1; i < mnum; i++)
		Add(src, i, 1, 0);
	for(int i = 1; i < mnum; i++)
		for(int j = 1; j < hnum; j++)
			Add(i, mnum + j - 1, 1, Get_dis(i, j));
	for(int j = 1; j < hnum; j++)
		Add(mnum + j - 1, sk, 1, 0);
}

bool SPFA()
{
	memset(vis, false, sizeof(vis));
	for(int i = 0; i < MAX; i++)
		dis[i] = INF;
	memset(pre, -1, sizeof(pre));
	queue <int> q;
	q.push(src);
	vis[src] = true;
	dis[src] = 0;
	while(!q.empty())
	{
		int u = q.front();
		q.pop();
		vis[u] = false;
		for(int i = head[u]; i != -1; i = e[i].nxt)
		{
			int v = e[i].to;
			int cap = e[i].cap;
			int cost = e[i].cost;
			if(cap > 0 && dis[v] > dis[u] + cost)
			{
				dis[v] = dis[u] + cost;
				pre[v] = i;
				if(!vis[v])
				{
					vis[v] = true;
					q.push(v);
				}
			}
		}
	}
	return dis[sk] != INF;
}

void Augment()
{
	int mi = INF;
	while(SPFA())
	{
		for(int i = sk; i != src; i = e[pre[i] ^ 1].to)
			mi = min(mi, e[pre[i]].cap);
		for(int i = sk; i != src; i = e[pre[i] ^ 1].to)
		{
			ans += mi * e[pre[i]].cost;
			e[pre[i]].cap -= mi;
			e[pre[i] ^ 1].cap += mi; 
		}
	}
}

int main()
{
	while(scanf("%d %d", &n, &m) && (n + m))
	{
		Init();
		Build_graph();
		Augment();
		printf("%d\n", ans);
	}
}


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