DFS-PATA-1004. Counting Leaves (30)
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题意:
给出N,M: 节点总个数,非叶节点个数
M行:根节点 根节点后的个数 节点编号
思路:
思路一:可以建树查询
思路二:用数组记录每个编号节点后有几个根节点,如果为空则层数+1;
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <vector>
#include <iostream>
using namespace std;
vector <int >adjlist[105];
int record [101]={0};
void dfs(int id ,int level)
{
if(adjlist[id].empty())
{
record[level]++;
return ;
}
for(vector<int>::iterator i=adjlist[id].begin();i!=adjlist[id].end();++i)
{
dfs(*i,level+1);
}
}
int main()
{
int n, k, x , y;
cin>>n>>k;
int cle=n-k; //记录下多少空点
for(int i=1;i<=k; i++)
{
int m;
cin>>x>>m;
while(m--)
{
cin>>y;
adjlist[x].push_back(y);
}
}
dfs(1,0);
printf("%d",record[0]);
int cnt = record[0];
for(int i=1;cnt<cle;++i){
printf(" %d",record[i]);
cnt += record[i]; //当前遍历到的以节点为根的空节点个数
}
printf("\n");
return 0;
}
#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <vector>#include <iostream>using namespace std;vector <int >adjlist[105];int record [101]={0};void dfs(int id ,int level){ if(adjlist[id].empty()) { record[level]++; return ; } for(vector<int>::iterator i=adjlist[id].begin();i!=adjlist[id].end();++i) { dfs(*i,level+1); }}int main(){ int n, k, x , y; cin>>n>>k; int cle=n-k; //记录下多少空点 for(int i=1;i<=k; i++) { int m; cin>>x>>m; while(m--) { cin>>y; adjlist[x].push_back(y); } } dfs(1,0);printf("%d",record[0]); int cnt = record[0];for(int i=1;cnt<cle;++i){printf(" %d",record[i]);cnt += record[i]; //当前遍历到的以节点为根的空节点个数}printf("\n");return 0;}
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