前言：
六度空间理论大家都知道，就是任何两个人之间只需要不超过6个人即可联系。
那么我们看看代码：
#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%dn",a)#define pfI(a) printf("%I64dn",a)#define pfs(a) printf("%sn",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m;#define Mod 10#define N 10010#define M 1000010struct Edge{//存储边的结构体int to;int next;};struct Node{//存储点的结构体int id;int step;};struct sqQueue{//队列的结构体Node *base;int front;int rear;};Edge edge[N];//int head[N];int vis[N];int cnt;sqQueue Q;void Init(){memset(head,-1,sizeof head);cnt=0;}void addEdge(int u,int v){//添加边edge[cnt].to = v;edge[cnt].next = head[u];head[u] = cnt++;}void InitQueue(sqQueue &Q){//初始化队列Q.base = new Node[M];if(!Q.base)exit(OVERFLOW);Q.front = Q.rear = 0;return;}int QueueLength(sqQueue Q){//求队列的长度return(Q.rear-Q.front+M)%M;}int EnQueue(sqQueue &Q,Node e){//入队if((Q.rear+1)%M == Q.front)return 0;Q.base[Q.rear] = e;Q.rear = (Q.rear+1)%M;return 1;}int DeQueue(sqQueue &Q){//出队if(Q.front == Q.rear)return 0;Q.front = (Q.front+1)%M;return 1;}Node getHead(sqQueue Q){//得到队列头部的元素return Q.base[Q.front];}void bfs(){//用bfs来搜索与初始点距离不超过7的点的个数占总数的百分比memset(vis,0,sizeof vis);InitQueue(Q);Node cur,next;cur.id = 0;cur.step = 0;EnQueue(Q,cur);vis[0] = 1;int num = 0;while(QueueLength(Q)){cur = getHead(Q);DeQueue(Q);if(cur.step>7){break;}num++;int s = cur.id;for(int p = head[s];p!=-1;p=edge[p].next){int v = edge[p].to;if(!vis[v]){next.id = v;next.step = cur.step + 1;vis[v] = 1;EnQueue(Q,next);}}}printf("%.4fn",num*1.0/n);}int main(){#ifndef ONLINE_JUDGE freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);#endif while(sfd(n,m)!=EOF){//n为顶点的数量，m为边的数量,点的编号从0-n-1 Init();//初始化 int u,v; for(int i=0;i<m;i++){ scanf("%d%d",&u,&v);//输入两个相邻的点 addEdge(u,v); addEdge(v,u);//无向边 } bfs(); }return 0;}
验证数据是in.txt文件，文件验证数据可以是：
10 9
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
后记：
*代码可以改进可以评论哦，谢谢！