POJ3697+BFS+hash存边
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1 /* 2 疾速优化+hash存边 3 题意:给定一个包含N(1 ≤ N ≤ 10,000)个顶点的无向完全图,图中的顶点从1到N依次标号。从这个图中去掉M(0 ≤ M ≤ 1,000,000)条边,求最后与顶点1联通的顶点的数目 4 思路(BFS):从顶点1开始不断扩展,广度优先搜索所有的与当前扩展点联通的顶点。开始每次都要判断所有的顶点是否与cur相连, 5 若相连则push,反之跳过。 6 */ 7 #include<stdio.h> 8 #include<string.h> 9 #include<stdlib.h> 10 #include<algorithm> 11 #include<iostream> 12 #include<queue> 13 #include<map> 14 #include<stack> 15 #include<set> 16 #include<math.h> 17 using namespace std; 18 typedef long long int64; 19 //typedef __int64 int64; 20 typedef pair<int64,int64> PII; 21 #define MP(a,b) make_pair((a),(b)) 22 const int maxn = 10005; 23 const int maxm = 1200005; 24 const int smod = 10091; 25 const int mod = 1000117; 26 const int inf = 0x7fffffff; 27 const double pi=acos(-1.0); 28 const double eps = 1e-8; 29 30 struct Edge{ 31 int u,v,next; 32 }edge[ maxm<<1 ]; 33 int cnt,myhash[ maxm<<1 ]; 34 queue<int>q; 35 int vis[ maxn ]; 36 37 void init(){ 38 cnt = 0; 39 while( !q.empty() ) 40 q.pop(); 41 memset( myhash,-1,sizeof( myhash ) ); 42 } 43 44 void addedge( int a,int b ){ 45 //if( a>b ) swap( a,b ); 46 int tt = (a*smod+b)%mod; 47 edge[ cnt ].u = a; 48 edge[ cnt ].v = b; 49 edge[ cnt ].next = myhash[ tt ]; 50 myhash[ tt ] = cnt++; 51 52 tt = (b*smod+a)%mod; 53 edge[ cnt ].u = a; 54 edge[ cnt ].v = b; 55 edge[ cnt ].next = myhash[ tt ]; 56 myhash[ tt ] = cnt++; 57 58 } 59 60 bool find( int u,int v ){ 61 //if( u>v ) swap( u,v ); 62 int tt = ( u*smod+v )%mod; 63 for( int i=myhash[tt];i!=-1;i=edge[i].next ){ 64 if( edge[ i ].u == u && edge[ i ].v == v ){ 65 return true; 66 } 67 } 68 return false; 69 } 70 71 int main(){ 72 int n,m; 73 int Case = 1; 74 while( scanf("%d%d",&n,&m)==2,n+m ){ 75 init(); 76 int x,y; 77 while( m-- ){ 78 scanf("%d%d",&x,&y); 79 addedge( x,y ); 80 addedge( y,x ); 81 } 82 int ans = 0; 83 cnt = 0; 84 q.push( 1 ); 85 for( int i=2;i<=n;i++ ){ 86 vis[ cnt++ ] = i; 87 } 88 while( !q.empty() ){ 89 int cur = q.front(); 90 q.pop(); 91 int tmp_cnt = 0; 92 for( int i=0;i<cnt;i++ ){ 93 if( !find( cur,vis[i] ) ){ 94 ans ++; 95 q.push( vis[i] ); 96 } 97 else vis[ tmp_cnt++ ] = vis[i]; 98 } 99 cnt = tmp_cnt; 100 } 101 printf("Case %d: %d\n",Case++,ans); 102 } 103 return 0; 104 }
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