二分图匹配(匈牙利算法) UVA 670 The dog task

  1 /*
  2     题意：bob按照指定顺序行走，他的狗可以在他到达下一个点之前到一个景点并及时返回，问狗最多能走多少个景点
  3     匈牙利算法：按照狗能否顺利到一个景点分为两个集合，套个模板
  4 */
  5 #include <cstdio>
  6 #include <algorithm>
  7 #include <cstring>
  8 #include <cmath>
  9 #include <vector>
 10 using namespace std;
 11 
 12 const int MAXN = 1e2 + 10;
 13 const int INF = 0x3f3f3f3f;
 14 struct P
 15 {
 16     int x, y;
 17     double nxt;
 18 }bob[MAXN], dog[MAXN];
 19 double d[MAXN][MAXN];
 20 bool vis[MAXN];
 21 int lk[MAXN];
 22 int lk2[MAXN];
 23 vector<int> G[MAXN];
 24 
 25 double get_dis(int x1, int y1, int x2, int y2)
 26 {
 27     return sqrt ((1.0) * ((x1-x2) * (x1-x2) + (y1-y2) * (y1-y2)));
 28 }
 29 
 30 bool DFS(int u)
 31 {
 32     for (int i=0; i<G[u].size (); ++i)
 33     {
 34         int v = G[u][i];
 35         if (!vis[v])
 36         {
 37             vis[v] = true;
 38             if (lk[v] == -1 || DFS (lk[v]))
 39             {
 40                 lk[v] = u;    lk2[u] = v;    return true;
 41             }
 42         }
 43     }
 44 
 45     return false;
 46 }
 47 
 48 void hungary(int n)
 49 {
 50     int res = 0;    memset (lk, -1, sizeof (lk));    memset (lk2, -1, sizeof (lk2));
 51     for (int i=1; i<n; ++i)
 52     {
 53         memset (vis, false, sizeof (vis));
 54         if (DFS (i))    res++;
 55     }
 56 
 57     printf ("%d\n", n + res);
 58     for (int i=1; i<n; ++i)
 59     {
 60         printf ("%d %d ", bob[i].x, bob[i].y);
 61         if (lk2[i] != -1)    printf ("%d %d ", dog[lk2[i]].x, dog[lk2[i]].y);
 62     }
 63     printf ("%d %d\n", bob[n].x, bob[n].y);
 64 }
 65 
 66 int main(void)        //UVA 670 The dog task
 67 {
 68 //    freopen ("UVA_670.in", "r", stdin);
 69 
 70     int t;    scanf ("%d", &t);
 71     while (t--)
 72     {
 73         int n, m;    scanf ("%d%d", &n, &m);
 74         for (int i=1; i<=n; ++i)
 75         {
 76             scanf ("%d%d", &bob[i].x, &bob[i].y);    G[i].clear ();
 77         }
 78         for (int i=1; i<=m; ++i)
 79         {
 80             scanf ("%d%d", &dog[i].x, &dog[i].y);
 81         }
 82 
 83         for (int i=1; i<=n; ++i)
 84         {
 85             for (int j=1; j<=m; ++j)
 86             {
 87                 d[i][j] = get_dis (bob[i].x, bob[i].y, dog[j].x, dog[j].y);
 88             }
 89         }
 90 
 91         for (int i=1; i<n; ++i)
 92         {
 93             bob[i].nxt = get_dis (bob[i].x, bob[i].y, bob[i+1].x, bob[i+1].y);
 94         }
 95 
 96         for (int i=1; i<n; ++i)
 97         {
 98             for (int j=1; j<=m; ++j)
 99             {
100                 if (d[i][j] + d[i+1][j] <= 2 * bob[i].nxt)    G[i].push_back (j);
101             }
102         }
103 
104         hungary (n);
105         if (t)    puts ("");
106     }
107 
108     return 0;
109 }