题目描述

注：题目可在牛客竞赛OJ找到并提交

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.

输入描述:

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.

输出描述:

For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.

​ 示例1

输入

复制;)

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

输出

复制;)

2
10
28

题目分析：

题目大意是给你一个二维网格，里面标出了人和房子的位置，问我们人都走到一个房子里面，最少需要走多少。其实就是很直接的告诉你，让你求一个人和房子的最小匹配。

那我们直接建图的时候对边权取反求最大匹配，最后的 ans 取反就是最小的匹配，是一道二分图带权匹配的模板题。

python code：

import sys

inf, _inf = float("inf"), float('-inf')
def km():
    def dfs(x):
        visx[x] = 1
        for i in range(1,H+1):
            if not visy[i] and lx[x] + ly[i] == g[x][i]:
                visy[i] = 1
                if link[i] == 0 or dfs(link[i]):
                    link[i] = x
                    return 1
        return 0
    
    for i in range(1,P+1):
        while True:
            visx,visy = [0]*(P+1), [0]*(H+1)
            if dfs(i): break
            d = inf
            for j in range(1,P+1):
                if visx[j]:
                    for k in range(1,H+1):
                        if not visy[k]: d = min(d, lx[j] + ly[k] - g[j][k])
            if d == inf: return 0
            for j in range(1,P+1):
                if visx[j]: lx[j] -= d
            for j in range(1,H+1):
                if visy[j]: ly[j] += d
    return 1

while True:
    n,m = map(int, sys.stdin.readline().split())
    P,H = 0,0  #人，房子的数量
    if n == m and n == 0: break
    p,h = [(0,0)],[(0,0)]  #人和房子的位置
    for i in range(1,n+1):
        s = '.'+sys.stdin.readline()
        for j in range(1,m+1):
            if s[j] == 'H':
                H += 1
                h.append((i,j))
            if s[j] == 'm':
                P += 1
                p.append((i,j))
    g = [[0]*(H+1) for _ in range(H+1)]
    for i in range(1, P+1):    #对每一个人和房子建图连边
        for j in range(1, H+1):
            g[i][j] = -(abs(p[i][0]-h[j][0]) + abs(p[i][1] - h[j][1]))

    lx,ly,link = [_inf]*(P+1),[0]*(H+1),[0]*(P+1)
    for i in range(1,P+1):   #对x集合的顶标进行预处理
        for j in range(1,H+1):
            if g[i][j] != 0:
                lx[i] = max(lx[i], g[i][j])
    if km() == 1:
        ans = 0
        for i in range(1,H+1):
            ans += g[link[i]][i]
        print(-ans)