poj 3020 Antenna Placement 解题报告

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int maxn = 500 + 10;
 7 
 8 int vis[maxn], match[maxn];
 9 int map[maxn][maxn], mark[maxn][maxn];
10 int cnt, maxmatch;     // cnt: 顶点数  maxmatch：最大匹配数
11 
12 int dx[] = {0, 0, 1, -1};
13 int dy[] = {1, -1, 0, 0};
14 
15 int dfs(int x)
16 {
17     for (int i = 1; i <= cnt-1; i++)
18     {
19         if (!vis[i] && map[x][i])
20         {
21             vis[i] = 1;
22             if (!match[i] || dfs(match[i]))
23             {
24                 match[i] = x;
25                 return 1;
26             }
27         }
28     }
29     return 0;
30 }
31 
32 void Hungary()
33 {
34     maxmatch = 0;
35     for (int i = 1; i <= cnt-1; i++)
36     {
37         memset(vis, 0, sizeof(vis));
38         maxmatch += dfs(i);
39     }
40 }
41 
42 int main()
43 {
44     int T, h, w;
45     while (scanf("%d", &T) != EOF)
46     {
47         while (T--)
48         {
49             scanf("%d%d", &h, &w);
50             char ch;
51             cnt = 1;
52             memset(mark, 0, sizeof(mark));
53             memset(map, 0, sizeof(map));
54             memset(match, 0, sizeof(match));
55 
56             for (int i = 0; i < h; i++)
57             {
58                 for (int j = 0; j < w; j++)
59                 {
60                     cin >> ch;
61                     if (ch == ‘*‘)
62                     {
63                         mark[i][j] = cnt;
64                         cnt++;
65                     }
66                 }
67             }
68             for (int i = 0; i < h; i++)
69             {
70                 for (int j = 0; j < w; j++)
71                 {
72                     if (mark[i][j])
73                     {
74                         int tmp = mark[i][j];     // 当前"*"编号
75                         for (int k = 0; k < 4; k++)   // 遍历它的四个方向
76                         {
77                             int ti = i + dx[k];
78                             int tj = j + dy[k];
79                             if (ti >= 0 && ti < 40 && tj >= 0 && tj < 10 && mark[ti][tj] > 0)
80                                 map[tmp][mark[ti][tj]] = 1;
81                             }
82                         }
83                     }
84                 }
85             }
86             Hungary();
87             printf("%d\n", cnt-1-maxmatch/2);
88         }
89     }
90     return 0;
91 }