hdu4460(BFS)

Friend Chains

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1860    Accepted Submission(s): 648

Problem Description

For a group of people, there is an idea that everyone is equals to or less than 6 steps away from any other person in the group, by way of introduction. So that a chain of "a friend of a friend" can be made to connect any 2 persons and it contains no more than 7 persons.
For example, if XXX is YYY’s friend and YYY is ZZZ’s friend, but XXX is not ZZZ's friend, then there is a friend chain of length 2 between XXX and ZZZ. The length of a friend chain is one less than the number of persons in the chain.
Note that if XXX is YYY’s friend, then YYY is XXX’s friend. Give the group of people and the friend relationship between them. You want to know the minimum value k, which for any two persons in the group, there is a friend chain connecting them and the chain's length is no more than k .

 

Input

There are multiple cases.
For each case, there is an integer N (2<= N <= 1000) which represents the number of people in the group.
Each of the next N lines contains a string which represents the name of one people. The string consists of alphabet letters and the length of it is no more than 10.
Then there is a number M (0<= M <= 10000) which represents the number of friend relationships in the group.
Each of the next M lines contains two names which are separated by a space ,and they are friends.
Input ends with N = 0.

 

Output

For each case, print the minimum value k in one line.
If the value of k is infinite, then print -1 instead.

 

Sample Input

3XXXYYYZZZ2XXX YYYYYY ZZZ0

 

Sample Output

2

 

Source

2012 Asia Hangzhou Regional Contest

        本题要求通过k朋友过渡认识有所人的最小的k.

        可以想到本题是求所有顶点间的最小距离的最大值。 2<=N<=1000,time limite=1000Ms,即说明一般情况下时间复杂度不应超过O(N^2)，故基本排除Floyd算法。我们知道BFS的时间复杂度为O(N)，我们可以利用变形的BFS求得单源最短路径，然后求1-n的所有顶点的单源最短路径，时间复杂度为O(N^2)。中间有些剪枝，用库函数的queue队列超时，得自己写数组模仿队列。还有可以将map容器写成线段树。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<map>//#include<queue>using namespace std;const int MAXN=1000+100;const int INF = 0x3f3f3f3f;int head[MAXN];int dis[MAXN];bool visited[MAXN];int qq[MAXN];struct node{int to;int next;}Edg[20000+100];int n,m,tot;map<string,int>g;void init(){tot=0;memset(head,-1,sizeof(head));}void add(int a,int b){Edg[tot].to=b;Edg[tot].next=head[a];head[a]=tot++;}int BFS(int s){//queue<int>qq;//qq.push(s);int front,rear;front=rear=0;qq[front++]=s;int now,i,to;for(i=1;i<=n;i++){if(i!=s)dis[i]=INF;else dis[i]=0;}visited[s]=true;while(rear<front){//now=qq.front();//qq.pop();now=qq[rear++];for(i=head[now];i!=-1;i=Edg[i].next){to=Edg[i].to;if(to==now||visited[to])continue;dis[to]=dis[now]+1;//qq.push(to);qq[front++]=to;visited[to]=true;}}int ret=-1;for(i=1;i<=n;i++){ret=max(ret,dis[i]);if(ret==INF)return ret;}return ret;}void Solve(){int i,ans=-1;for(i=1;i<=n;i++){memset(visited,0,sizeof(visited));ans=max(ans,BFS(i));if(ans==INF)break;}if(ans==INF)printf("-1\n");else printf("%d\n",ans);}int main(){int i;string st,ed;while(scanf("%d",&n),n){g.clear();for(i=1;i<=n;i++){cin>>st;g[st]=i;}scanf("%d",&m);init();for(i=1;i<=m;i++){cin>>st>>ed;add(g[st],g[ed]);add(g[ed],g[st]);}Solve();}return 0;}