hdu 4460 Friend Chains

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=4460

 

题目描述：

Friend Chains

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 489    Accepted Submission(s): 235

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

 

 

 

题意：给出一个无向图，若联通则输出任意一对点之间最短路径中的最大值，若不联通则输出-1。

 

题解：对每个点做n次bfs求最短路径，判断图的连通性，注意考虑-1的情况，避免WA。另外bfs的时候利用vector存储一个点的子节点来优化bfs一个父结点时寻找 有哪些子节点，这样就不用穷举所有点判断此点是不是这个父节点的子节点。不然会TLE。

 

代码：

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<queue>#include<map>#include<iostream>#include<string>#include<vector>using namespace std;map<string,int> NameMap;bool Graph[1000+5][1000+5]={false};bool Vis[1000+5]={false};int KLen[1000+5]={0};int N=0,M=0;string Str,Str1;int k=0,K=0;vector<int> vec[1000+5];/*BFS the whole graph*/int BFS(int cur){queue<int> Queues;Queues.push(cur);Vis[cur]=true;while(!Queues.empty()){int popcur=Queues.front();Queues.pop();int i=0;int count=vec[popcur].size();//optimizationfor(i=0;i<=count-1;i++){int j=vec[popcur][i];if(Graph[popcur][j]&&!Vis[j]){KLen[j]=KLen[popcur]+1;Queues.push(j);Vis[j]=true;if(KLen[j]>k){k=KLen[j];}}}}return(0);}/*for test*/int test(){return(0);}/*main process*/int MainProc(){while(scanf("%d",&N)!=EOF&&N>0){memset(Graph,false,sizeof(Graph));memset(Vis,false,sizeof(Vis));memset(KLen,0,sizeof(KLen));NameMap.clear();k=0;K=0;int i=0;for(i=0;i<=N-1;i++){cin>>Str;NameMap[Str]=i;//from 0 to N-1vec[i].clear();}scanf("%d",&M);for(i=0;i<=M-1;i++){cin>>Str>>Str1;Graph[NameMap[Str]][NameMap[Str1]]=true;Graph[NameMap[Str1]][NameMap[Str]]=true;vec[NameMap[Str]].push_back(NameMap[Str1]);vec[NameMap[Str1]].push_back(NameMap[Str]);}for(i=0;i<=N-1;i++)//n count of bfs to the graph{k=0;memset(Vis,false,sizeof(Vis));memset(KLen,0,sizeof(KLen));BFS(i);if(k>0&&K>=0&&K<k){K=k;}if(k<=0){K=-1;}}printf("%d\n",K);}return(0);}int main(){MainProc();return(0);}