PAT (Advanced Level) Practise - 1004. Counting Leaves (30)
2017-11-04 22:39
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1004. Counting Leaves (30)
时间限制400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input
Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tree, and M (< N), the number of non-leaf nodes. Then M lines follow, each in the format:
ID K ID[1] ID[2] ... ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.
Output
For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end
of each line.
The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output "0 1" in a line.
Sample Input
2 1 01 1 02
Sample Output
0 1
题意:
给你一棵树,让你求出每层中有几个节点没有子节点。
我没有用DFS来做,我直接模拟了。
#include <iostream> #include <algorithm> #include <stdio.h> #include <string.h> #include <math.h> #include <map> using namespace std; #define LL long long const int maxn = 111; const int inf = 0x3f3f3f3f; int mp[maxn][maxn]; int flag[maxn]; int cnt=0; int main() { int n,m; memset(flag,-1,sizeof flag); scanf("%d %d",&n,&m); for(int i=1;i<=m;i++){ int fa,k,son; scanf("%d %d",&fa,&k); for(int i=1;i<=k;i++){ scanf("%d",&son); mp[fa][son]=1; flag[fa]=1; if(flag[son]==-1){ flag[son]=0; } } } int num[maxn]; num[++cnt]=1; int fnum[maxn]; int pp[maxn]; int pcnt=0; int *r=num,*p=fnum; while(cnt!=0){ int fcnt=0; int ans=0; for(int i=1;i<=cnt;i++){ int m=1; int u=r[i]; for(int j=1;j<=99;j++){ if(mp[u][j]){ p[++fcnt]=j; m=0; } } if(m==1) ans++; } cnt=fcnt; swap(r, p); pp[++pcnt]=ans; } for(int i=1;i<=pcnt;i++){ if(i-1) printf(" "); printf("%d",pp[i]); } printf("\n"); }
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