【PAT】1094. The Largest Generation
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考查点：BFS或DFS，求树的最大节点层
思路：BFS代码长点，DFS较短，BFS只需设置个变量last来记录每一层的最后一个节点；DFS只需两个参数，一个节点一个层数，每次调用都要更新当前层的节点数，可以用哈希数组保存层数的节点值
提交情况：一开始输入时候%d写错成%k调了半天，之后因为找根节点时候没把0排除，第一次提交没考虑只有一个根节点的情况之后ac
BFS的代码：
#define LOCAL#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#define FOR(i, x, y) for(int i = x; i < y; i++)#define rFOR(i, x, y) for(int i = x; i > = y; i--)#define MAXN 110#define oo 0x3f3f3f3fusing namespace std;int n;int h[MAXN];vector<int> child[MAXN];int ans;int alevel;void BFS(int root){ queue<int> q; int level=1; int cnt=1; int first=0,last=root; q.push(root); while(!q.empty()){ int t=q.front(); q.pop(); FOR(i,0,child[t].size()) { q.push(child[t][i]); } if(t==last){ cnt=q.size(); last=q.back(); level++; } if(cnt>ans){ ans=cnt;alevel=level; } }}int main(){ #ifdef LOCAL freopen("data.in","r",stdin); freopen("data.out","w",stdout); #endif // LOCAL int lv,id,k,x; scanf("%d%d",&n,&lv); FOR(i,0,lv) { scanf("%d%d",&id,&k); FOR(j,0,k) { scanf("%d",&x); child[id].push_back(x); h[x]=1; } } int root; FOR(i,1,n+1){ if(h[i]==0) { root=i;break; } } BFS(root); if(n!=1){ printf("%d %d",ans,alevel); }else printf("1 1"); return 0;}DFS的代码：#define LOCAL#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#define FOR(i, x, y) for(int i = x; i < y; i++)#define rFOR(i, x, y) for(int i = x; i > = y; i--)#define MAXN 110#define oo 0x3f3f3f3fusing namespace std;vector<int> child[MAXN];int ha[MAXN];void DFS(int index,int level){ ha[level]++; FOR(j,0,child[index].size()) { DFS(child[index][j],level+1); }}int main(){ #ifdef LOCAL freopen("data.in","r",stdin); freopen("data.out","w",stdout); #endif // LOCAL int n,m,parent,k,ch; scanf("%d%d",&n,&m); FOR(i,0,m) { scanf("%d%d",&parent,&k); FOR(j,0,k) { scanf("%d",&ch); child[parent].push_back(ch); } } DFS(1,1); int maxLevel=-1,maxValue=0; FOR(i,1,MAXN) { if(ha[i]>maxValue){ maxValue=ha[i]; maxLevel=i; } } printf("%d %d\n",maxValue,maxLevel); return 0;}0 0
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