There is a tree having N vertices. In the tree there are K monkeys (K <= N). A vertex can be occupied by at most one monkey. They want to remove some edges and leave minimum edges, but each monkey must be connected to at least one other monkey through the remaining edges.
Print the minimum possible number of remaining edges.
InputThe first line contains an integer T (1 <= T <= 100), the number of test cases.
Each test case begins with a line containing two integers N and K (2 <= K <= N <= 100000). The second line contains N-1 space-separated integers a 1 ,a 2 ,…,a N−1 a1,a2,…,aN−1
, it means that there is an edge between vertex a i ai
and vertex i+1 (1 <= a i ai
<= i).
OutputFor each test case, print the minimum possible number of remaining edges.Sample Input
2 4 4 1 2 3 4 3 1 1 1
Sample Output
2 2
题意：给定一棵树，有K个猴子，现在让你把猴子放到节点上（一个节点最多一个猴子），求最少留多少边，使得每个连通块的猴子数不为1 。
思路：尽量把两个猴子用一条边连接起来。所以我们先贪心，有多少个“边匹配”，假设最大值为cnt。然后如果够，那么答案是（K+1）/2；如果不够，则加边即可，答案是cnt+(K-cnt*2)。
求最大“边匹配”的方法是贪心，从下想上贪，因为一个点最大贡献到一条边，所以把u和没有被匹配的fa[u]匹配，其效果不会比fa[u]和fa[fa[u]]匹配差。
不用输入优化会T。
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
using namespace std;
const int maxn=1000010;
int vis[maxn],fa[maxn],cnt,ans;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline void read(int &sum){
char ch=nc(); sum=0;
while(!(ch>=‘0‘&&ch<=‘9‘))ch=nc();
while(ch>=‘0‘&&ch<=‘9‘) sum=sum*10+ch-48,ch=nc();
}
int main()
{
int T,N,K;
read(T);
while(T--){
read(N);read(K); ans=cnt=0;
rep(i,1,N) vis[i]=0;
rep(i,2,N) read(fa[i]);
for(int i=N;i>=2;i--)
if(!vis[i]&&!vis[fa[i]]) ans++,vis[fa[i]]=1;
if(ans*2>=K) printf("%d\n",(K+1)/2);
else printf("%d\n",ans+(K-ans*2));
}
return 0;
}
原文地址：https://www.cnblogs.com/hua-dong/p/9822997.html