Give you a tree, can you draw the tree with minimum strokes without overlapping? Noted that it is ok if two strokes intersect at one point. Here we define a tree as a connected undirected graph with N points and N-1 edges.
Input
The input data has several test cases. The first line contains a positive integer T (T<=20), indicates the number of test cases.
For each case:
The first line contains an integers N (2<=N<=10^5), indicates the number of points on the tree numbered from 1 to N.
Then follows N-1 lines, each line contains two integers Xi, Yi means an edge connected Xi and Yi (1<=Xi, Yi<=N).
Output
For each test case, you should output one line with a number K means the minimum strokes to draw the tree.
Sample Input
2
2
1 2
5
1 2
1 5
2 3
2 4
Sample Output
1
2
题意:
已知一颗n个结点,n-1条边的树,树上边为无向边,求最少需要花几次才能把这棵树画出来,走过的
路径不能重复
思路:
从一个奇度顶点出发,到另一个奇度顶点,为一条路径
需要画的次数ans = 奇度顶点个数/2
代码:
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1e5+5;
int Deg[N],n;
int main(){
int T;
scanf("%d",&T);
while(T--){
memset(Deg,0,sizeof(Deg));
scanf("%d",&n);
for(int i=1;i<n;i++){
int u,v;
scanf("%d%d",&u,&v);
Deg[u]++;
Deg[v]++;
}
int ans=0;
for(int i=1;i<=n;i++)
ans+=(Deg[i]&1);
printf("%d\n",ans/2);
}
return 0;
}