hdu 5424 回溯+并查集判断连通性

题意：给定一个n个点n条边的无向图，判断是否存在哈密顿路径。

思路：先用并查集判断图是否连通，然后从度数最小的点开始回溯，看是否能找到一条哈密顿路径。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 #include <set>
  5 using namespace std;
  6
  7 const int INF = 999999;
  8 const int N = 1001;
  9 int f[N];
 10 int d[N];
 11 int head[N];
 12 bool visit[N];
 13 int n, e;
 14
 15 void init()
 16 {
 17     e = 0;
 18     memset( d, 0, sizeof(d) );
 19     memset( head, -1, sizeof(head) );
 20     memset( visit, 0, sizeof(visit) );
 21     for ( int i = 1; i < N; i++ ) f[i] = i;
 22 }
 23
 24 struct Edge
 25 {
 26     int v, next;
 27 } edge[N << 1];
 28
 29 void addEdge( int u, int v )
 30 {
 31     edge[e].v = v;
 32     edge[e].next = head[u];
 33     head[u] = e++;
 34 }
 35
 36 int findf( int x )
 37 {
 38     if ( f[x] != x ) f[x] = findf( f[x] );
 39     return f[x];
 40 }
 41
 42 bool union_set( int x, int y )
 43 {
 44     x = findf(x), y = findf(y);
 45     if ( x == y )
 46     {
 47         return false;
 48     }
 49     else
 50     {
 51         f[x] = y;
 52         return true;
 53     }
 54 }
 55
 56 bool dfs( int u, int cnt )
 57 {
 58     if ( cnt == n ) return true;
 59     for ( int i = head[u]; i != -1; i = edge[i].next )
 60     {
 61         int v = edge[i].v;
 62         if ( !visit[v] )
 63         {
 64             visit[v] = 1;
 65             if ( dfs( v, cnt + 1 ) ) return true;
 66             visit[v] = 0;
 67         }
 68     }
 69     return false;
 70 }
 71
 72 int main ()
 73 {
 74     while ( scanf("%d", &n) != EOF )
 75     {
 76         init();
 77         int c = 0;
 78         for ( int i = 1; i <= n; i++ )
 79         {
 80             int u, v;
 81             scanf("%d%d", &u, &v);
 82             addEdge( u, v );
 83             addEdge( v, u );
 84             d[u]++, d[v]++;
 85             if ( union_set( u, v ) ) c++;
 86         }
 87         if ( c != n - 1 )
 88         {
 89             puts("NO");
 90             continue;
 91         }
 92         d[0] = INF;
 93         int k = 0;
 94         for ( int i = 1; i <= n; i++ )
 95         {
 96             if ( d[i] < d[k] )
 97             {
 98                 k = i;
 99             }
100         }
101         visit[k] = 1;
102         if ( dfs( k, 1 ) )
103         {
104             puts("YES");
105         }
106         else
107         {
108             puts("NO");
109         }
110     }
111     return 0;
112 }