二分图最大匹配(匈牙利算法) UVA 670 The dog task

题目传送门

 /*
     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
     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
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <vector>
 using namespace std;

 const int MAXN = 1e2 + ;
 const int INF = 0x3f3f3f3f;
 struct P
 {
     int x, y;
     double nxt;
 }bob[MAXN], dog[MAXN];
 double d[MAXN][MAXN];
 bool vis[MAXN];
 int lk[MAXN];
 int lk2[MAXN];
 vector<int> G[MAXN];

 double get_dis(int x1, int y1, int x2, int y2)
 {
     return sqrt ((1.0) * ((x1-x2) * (x1-x2) + (y1-y2) * (y1-y2)));
 }

 bool DFS(int u)
 {
     for (int i=; i<G[u].size (); ++i)
     {
         int v = G[u][i];
         if (!vis[v])
         {
             vis[v] = true;
             if (lk[v] == - || DFS (lk[v]))
             {
                 lk[v] = u;    lk2[u] = v;    return true;
             }
         }
     }

     return false;
 }

 void hungary(int n)
 {
     int res = ;    memset (lk, -, sizeof (lk));    memset (lk2, -, sizeof (lk2));
     for (int i=; i<n; ++i)
     {
         memset (vis, false, sizeof (vis));
         if (DFS (i))    res++;
     }

     printf ("%d\n", n + res);
     for (int i=; i<n; ++i)
     {
         printf ("%d %d ", bob[i].x, bob[i].y);
         if (lk2[i] != -)    printf ("%d %d ", dog[lk2[i]].x, dog[lk2[i]].y);
     }
     printf ("%d %d\n", bob[n].x, bob[n].y);
 }

 int main(void)        //UVA 670 The dog task
 {
 //    freopen ("UVA_670.in", "r", stdin);

     int t;    scanf ("%d", &t);
     while (t--)
     {
         int n, m;    scanf ("%d%d", &n, &m);
         for (int i=; i<=n; ++i)
         {
             scanf ("%d%d", &bob[i].x, &bob[i].y);    G[i].clear ();
         }
         for (int i=; i<=m; ++i)
         {
             scanf ("%d%d", &dog[i].x, &dog[i].y);
         }

         for (int i=; i<=n; ++i)
         {
             for (int j=; j<=m; ++j)
             {
                 d[i][j] = get_dis (bob[i].x, bob[i].y, dog[j].x, dog[j].y);
             }
         }

         for (int i=; i<n; ++i)
         {
             bob[i].nxt = get_dis (bob[i].x, bob[i].y, bob[i+].x, bob[i+].y);
         }

         for (int i=; i<n; ++i)
         {
             for (int j=; j<=m; ++j)
             {
                 if (d[i][j] + d[i+][j] <=  * bob[i].nxt)    G[i].push_back (j);
             }
         }

         hungary (n);
         if (t)    puts ("");
     }

     return ;
 }