集训队专题（7）1006 Going Home

Going Home

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3891    Accepted Submission(s): 1991

Problem 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

 

此题是一个基础的最小费用最大流问题，最小费用最大流问题的解法就是，先找一个最小费用可行流，再找出关于该可行流的最小费用增广链，沿此链调整流量，则得到一个新的流量增大了的最小费用流，然后对新的最小费用流重复上述方法，一直调整到网络的最大流出现为止，便得到了所考虑网络的最小费用最大流。

#include <cstdio>
#include <cstring>
#include <queue>
#include <cstdlib>
#include <algorithm>
#define MAXN 200+10
#define MAXM 80000+100
#define INF 0x3f3f3f3f
using namespace std;
struct Edge
{
    int from, to, cap, flow, cost, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int pre[MAXN], dist[MAXN];
bool vis[MAXN];
int N, M;
int cost, flow;
int sink, source;//超级源点 超级汇点
void init()
{
    edgenum = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w, int c)
{
    Edge E1 = {u, v, w, 0, c, head[u]};
    edge[edgenum] = E1;
    head[u] = edgenum++;
    Edge E2 = {v, u, 0, 0, -c, head[v]};
    edge[edgenum] = E2;
    head[v] = edgenum++;
}
int dis(int x1, int y1, int x2, int y2)
{
    return abs(x1 - x2) + abs(y1 - y2);
}
struct Node
{
    int x, y;
};
Node m[110], H[110];//存储字符坐标
int m_cnt;//m字符计数器
int H_cnt;//H字符计数器
void getMap()
{
    m_cnt = H_cnt = 0;
    char str[110][110];
    for(int i = 0; i < N; i++)
    {
        scanf("%s", str[i]);
        for(int j = 0; j < M; j++)
        {
            if(str[i][j] == 'm')
            {
                ++m_cnt;
                m[m_cnt].x = i;
                m[m_cnt].y = j;
            }
            if(str[i][j] == 'H')
            {
                ++H_cnt;
                H[H_cnt].x = i;
                H[H_cnt].y = j;
            }
        }
    }
    int k = m_cnt;//人数 或者 房子数
    sink = 0;
    source = 2*k+1;
    for(int i = 1; i <= k; i++)
    {
        addEdge(sink, i, 1, 0);
        addEdge(i + k, source, 1, 0);
        for(int j = 1; j <= k; j++)
        {
            int d = dis(H[i].x, H[i].y, m[j].x, m[j].y);
            addEdge(i, j + k, 1, d);
        }
    }
}
bool SPFA(int s, int t)//寻找花销最少的路径
{
    queue<int> Q;
    memset(dist, INF, sizeof(dist));
    memset(vis, false, sizeof(vis));
    memset(pre, -1, sizeof(pre));
    dist[s] = 0;
    vis[s] = true;
    Q.push(s);
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            Edge E = edge[i];
            if(dist[E.to] > dist[u] + E.cost && E.cap > E.flow)//可以松弛 且 没有满流
            {
                dist[E.to] = dist[u] + E.cost;
                pre[E.to] = i;//记录前驱边 的编号
                if(!vis[E.to])
                {
                    vis[E.to] = true;
                    Q.push(E.to);
                }
            }
        }
    }
    return pre[t] != -1;//可达返回true
}
void MCMF(int s, int t)
{
    flow = 0;//总流量
    cost = 0;//总费用
    while(SPFA(s, t))//每次寻找花销最小的路径
    {
        int Min = INF;
        //通过反向弧 在源点到汇点的最少花费路径 找最小增广流
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            Edge E = edge[i];
            Min = min(Min, E.cap - E.flow);
        }
        //增广
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost * Min;//增广流的花销
        }
        flow += Min;//流量累加
    }
}
int main()
{
    while(scanf("%d%d", &N, &M), N||M)
    {
        init();
        getMap();
        
        MCMF(sink, source);
        printf("%d\n", cost);
    }
    return 0;
}