POJ3697+BFS+hash存边

/*
疾速优化+hash存边
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
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，
  若相连则push，反之跳过。
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<math.h>
using namespace std;
typedef long long int64;
//typedef __int64 int64;
typedef pair<int64,int64> PII;
#define MP(a,b) make_pair((a),(b)) 
const int maxn = 10005;
const int maxm = 1200005;
const int smod = 10091;
const int mod = 1000117;
const int inf = 0x7fffffff;
const double pi=acos(-1.0);
const double eps = 1e-8;

struct Edge{
    int u,v,next;
}edge[ maxm<<1 ];
int cnt,myhash[ maxm<<1 ];
queue<int>q;
int vis[ maxn ];

void init(){
    cnt = 0;
    while( !q.empty() )
        q.pop();
    memset( myhash,-1,sizeof( myhash ) );
}

void addedge( int a,int b ){
    //if( a>b ) swap( a,b );
    int tt = (a*smod+b)%mod;
    edge[ cnt ].u = a;
    edge[ cnt ].v = b;
    edge[ cnt ].next = myhash[ tt ];
    myhash[ tt ] = cnt++;
    
    tt = (b*smod+a)%mod;
    edge[ cnt ].u = a;
    edge[ cnt ].v = b;
    edge[ cnt ].next = myhash[ tt ];
    myhash[ tt ] = cnt++;
    
}

bool find( int u,int v ){
    //if( u>v ) swap( u,v );
    int tt = ( u*smod+v )%mod;
    for( int i=myhash[tt];i!=-1;i=edge[i].next ){
        if( edge[ i ].u == u && edge[ i ].v == v ){
            return true;
        }
    }
    return false;
}

int main(){
    int n,m;
    int Case = 1;
    while( scanf("%d%d",&n,&m)==2,n+m ){
        init();
        int x,y;
        while( m-- ){
            scanf("%d%d",&x,&y);
            addedge( x,y );
            addedge( y,x );
        }
        int ans = 0;
        cnt = 0;
        q.push( 1 );
        for( int i=2;i<=n;i++ ){
            vis[ cnt++ ] = i;
        }
        while( !q.empty() ){
            int cur = q.front();
            q.pop();
            int tmp_cnt = 0;
            for( int i=0;i<cnt;i++ ){
                if( !find( cur,vis[i] ) ){
                    ans ++;
                    q.push( vis[i] );
                }
                else  vis[ tmp_cnt++ ] = vis[i];
            }
            cnt = tmp_cnt;
        }
        printf("Case %d: %d
",Case++,ans);
    }
    return 0;
}