解法一:动态规划
#includeusing namespace std; vector tree[20005];//存储整棵树 int n,m,a; int dp1[20005],dp2[20005]; void DP(int v){//深度优先遍历 int max1=0,max2=0; for(int i:tree[v]){ DP(i); if(dp1[i]+1>max1){//计算到根结点v到叶子结点的最远距离 max2=max1; max1=dp1[i]+1; }else if(dp1[i]+1>max2)//计算到根结点v到叶子结点的第二远距离 max2=dp1[i]+1; } dp1[v]=max1; dp2[v]=max1+max2; } int main(){ scanf("%d%d",&n,&m); for(int i=2;i<=n+m;++i){ scanf("%d",&a); tree[a].push_back(i); } DP(1); printf("%d",*max_element(dp2+1,dp2+n+m+1));//dp2中最大值即为所求 return 0; }
解法二:DFS
#includeusing namespace std; vector tree[20005];//存储整棵树 int n,m,a,ansV=-1,ansLevel=0;//ansV表示距离最远的结点编号,ansLevel表示最远距离 bool visit[20005];//标记每个节点是否已被访问 void DFS(int v,int level){//深度优先遍历 visit[v]=true; if(level>ansLevel){ ansLevel=level; ansV=v; } for(int i:tree[v]) if(!visit[i]) DFS(i,level+1); } int main(){ scanf("%d%d",&n,&m); for(int i=2;i<=n+m;++i){ scanf("%d",&a); tree[a].push_back(i); tree[i].push_back(a);//将树的边存储为无向边 } DFS(1,0);//第一次深度优先遍历 memset(visit,0,sizeof(visit)); DFS(ansV,0);//二次遍历 printf("%d",ansLevel); return 0; }



