POJ 3310 Caterpillar

Description:

　An undirected graph is called a caterpillar if it is connected, has no cycles, and there is a path in the graph where every node is either on this path or a neighbor of a node on the path. This path is called the spine of the caterpillar and the spine may not be unique. You are simply going to check graphs to see if they are caterpillars.
　For example, the left graph below is not a caterpillar, but the right graph is. One possible spine isshown by dots.

Input:

　There will be multiple test cases. Each test case starts with a line containing n indicating the number of nodes, numbered 1 through n (a value of n = 0 indicates end-of-input). The next line will contain an integer e indicating the number of edges. Starting on the following line will be e pairs n1 n2 indicating an undirected edge between nodes n1 and n1. This information may span multiple lines. You may assume that n ≤ 100 and e ≤ 300. Do not assume that the graphs in the test cases are connected or acyclic.

Output:

　For each test case generate one line of output. This line should either be
“Graph g is a caterpillar.”
or
“Graph g is not a caterpillar.”
as appropriate, where g is the number of the graph, starting at 1.

Sample Input:

22
21
1 2 2 3 2 4 2 5 2 6 6 7 6 10 10 8 9 10 10 12 11 12 12 13 12 17
18 17 15 17 15 14 16 15 17 20 20 21 20 22 20 19
16
15
1 2 2 3 5 2 4 2 2 6 6 7 6 8 6 9 9 10 10 12 10 11 10 14 10 13 13 16 13 15
0

Sample Output:

Graph 1 is not a caterpillar.
Graph 2 is a caterpillar.

题解：
　读完题目我们知道，题目要求就是找一条路径，满足对于任意路径外的点都可以在路径上找到一个点到它的距离为1。因为路径为1，所以那条路径肯定是树的直径。
　又因为n<=100,所以这道题可以直接n3找路径，然后直接将树的直径上的点存到数组里，然后n2判断距离是否为1
　还要注意题目要求不能有环，所以一开始还要判环，可以用dfs,也可以用并查集，我用的是并查集

#include<cstdio>#include<cmath>#include<ctime>#include<iomanip>#include<iostream>#include<cstring>#include<string>#include<cctype>#include<algorithm>#define N 105#define M 305#define INF 100000000using namespace std;int n,m,x,y,s,t,num,index,max_dis,tail;int p[105],fa[105],f[105][105];bool flag=false,bz=false,in[105];inline int Readint(){    int i=0,f=1;char ch;    for(ch=getchar();(ch<'0'||ch>'9')&&ch!='-';ch=getchar());    if(ch=='-') f=-1,ch=getchar();    for(;ch>='0'&&ch<='9';ch=getchar()) i=(i<<1)+(i<<3)+ch-'0';    return i*f;}inline int find(int x){    if(x==fa[x]) return x;    fa[x]=find(fa[x]);    return fa[x];}inline bool judge(){    int cnt=0;    for(int i=1;i<=n;i++){        if(fa[find(i)]==i)cnt++;    }    return cnt==1;}inline void pre(){    num=max_dis=tail=0;    bz=flag=false;    memset(in,0,sizeof(in));    memset(p,0,sizeof(p));    for(int i=1;i<=n;i++) fa[i]=i;    for(int i=1;i<=n;i++)      for(int j=1;j<=n;j++)        f[i][j]=INF;}int main(){//  freopen("lx.in","r",stdin);    while(scanf("%d",&n) && n!=0){        pre();        m=Readint();        for(int i=1;i<=m;i++){            x=Readint(),y=Readint();            f[x][y]=f[y][x]=1;            if(find(x)!=find(y)) fa[find(x)]=find(y);//判环            else flag=true;        }        if(m!=n-1){            cout<<"Graph "<<++index<<" is not a caterpillar."<<endl;            continue;        }        if(flag || !judge()) {            cout<<"Graph "<<++index<<" is not a caterpillar."<<endl;            continue;        }        for(int i=1;i<=n;i++) f[i][i]=0;        for(int k=1;k<=n;k++)          for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)              f[i][j]=min(f[i][j],f[i][k]+f[k][j]);        for(int i=1;i<=n;i++)          for(int j=1;j<=n;j++)            if(f[i][j]>max_dis && f[i][j]!=INF){                max_dis=f[i][j];                s=i,t=j;            }//找树的直径        p[++tail]=s; p[++tail]=t;        in[s]=in[t]=true;        for(int i=1;i<=n;i++)          if((f[s][i]+f[i][t])==f[s][t] && i!=s && i!=t)             p[++tail]=i,in[i]=true;//把树的直径上的点都存在p[]里        for(int i=1;i<=n;i++){            for(int j=1;j<=tail;j++){                if(f[i][p[j]]==1 && i!=s && i!=t && !in[i]){                    bz=true;                    num++;                }            }            if(bz==true) continue;        }         if(num==n-tail) cout<<"Graph "<<++index<<" is a caterpillar."<<endl;        else cout<<"Graph "<<++index<<" is not a caterpillar."<<endl;    }    return 0;}

最后还要注意：
在判环和判是否为一棵树时，一定要等这组数据读完再推出，否则下次会一直读这组数据剩下的数，我就是这样WA了一上午QAQ