最大流增广路(KM算法) HDOJ 1533 Going Home

题目传送门

 1 /*
 2     最小费用流：KM算法是求最大流，只要w = -w就可以了，很经典的方法
 3 */
 4 #include <cstdio>
 5 #include <cmath>
 6 #include <algorithm>
 7 #include <cstring>
 8 using namespace std;
 9
10 const int MAXN = 1e2 + 10;
11 const int INF = 0x3f3f3f3f;
12 int x[MAXN], y[MAXN];
13 int w[MAXN][MAXN];
14 int visx[MAXN], visy[MAXN];
15 int ly[MAXN];
16 int mx[MAXN], my[MAXN];
17 int hx[MAXN], hy[MAXN];
18 char maze[MAXN][MAXN];
19 int n, m, un, vn, d;
20
21 bool DFS(int u) {
22     visx[u] = true;
23     for (int i=1; i<=un; ++i)   {
24         if (!visy[i] && x[u] + y[i] == w[u][i]) {
25             visy[i] = true;
26             if (ly[i] == -1 || DFS (ly[i])) {
27                 ly[i] = u;  return true;
28             }
29         }
30         else if (x[u] + y[i] > w[u][i]) d = min (d, x[u] + y[i] - w[u][i]);
31     }
32     return false;
33 }
34
35 int KM(void)    {
36     for (int i=1; i<=un; ++i)   {
37         x[i] = -INF;
38         for (int j=1; j<=vn; ++j)   {
39             x[i] = max (x[i], w[i][j]);
40         }
41     }
42
43     memset (ly, -1, sizeof (ly));
44     memset (y, 0, sizeof (y));
45     for (int i=1; i<=un; ++i)    {
46         while (true)    {
47             memset (visx, false, sizeof (visx));
48             memset (visy, false, sizeof (visy));
49             d = INF;
50             if (DFS (i))    break;
51             for (int i=1; i<=un; ++i)   {
52                 if (visx[i])    x[i] -= d;
53             }
54             for (int j=1; j<=vn; ++j)   {
55                 if (visy[j])    y[j] += d;
56             }
57         }
58     }
59
60     int res = 0;
61     for (int i=1; i<=un; ++i)   {
62         res += x[i] + y[i];
63     }
64
65     return res;
66 }
67
68 int main(void)  {       //HDOJ 1533 Going Home
69     //freopen ("HDOJ_1533.in", "r", stdin);
70
71     while (scanf ("%d%d", &n, &m) == 2) {
72         if (!n && !m)   break;
73         for (int i=1; i<=n; ++i)    {
74             scanf ("%s", maze[i] + 1);
75         }
76         un = vn = 0;
77         for (int i=1; i<=n; ++i)    {
78             for (int j=1; j<=m; ++j)    {
79                 if (maze[i][j] == ‘m‘)  mx[++un] = i, my[un] = j;
80                 else if (maze[i][j] == ‘H‘) hx[++vn] = i, hy[vn] = j;
81             }
82         }
83         for (int i=1; i<=un; ++i)   {
84             for (int j=1; j<=vn; ++j)   {
85                 w[i][j] = -(abs (mx[i] - hx[j]) + abs (my[i] - hy[j]));
86             }
87         }
88         printf ("%d\n", -KM ());
89     }
90
91     return 0;
92 }