SCU 4313 把一棵树切成每段K个点 (n%k)剩下的点不管

题目链接：http://cstest.scu.edu.cn/soj/problem.action?id=4313

判断是不是存在拆图得到新连通分支的点个数是K的倍数

注意一个点所连的边只能被切一条

#include<stdio.h>
#include<string.h>

#define N 200001

struct node{
	int f,t,fn,tn,nex;
}edge[N];
int edgenum, head[N];
void addedge(int u, int v){
	node E={u,v,0,0,head[u]};
	edge[edgenum] = E;
	head[u] = edgenum++;
}
int n,K,m; //n个点 m条边
int du[N],dp[N];
bool vis[N], use[N];
int ans ;
void dfs(int x){
	for(int i=head[x]; i!=-1; i = edge[i].nex){
		int v = edge[i].t;
		if(!vis[v]){
			vis[v] = 1;
			dfs(v);
			dp[x] += dp[v];
			edge[i].tn = dp[v];
			edge[i].fn = n - edge[i].tn;
			if(edge[i].tn %K==0 && !use[v] )ans++, use[v] = 1;

			if(edge[i].fn %K==0 && !use[x] )ans++, use[x] = 1;
		}
	}
}
int main(){
	int T,i,j;scanf("%d",&T);
	while(T--){
		scanf("%d %d",&n,&K);


		memset(head,-1,sizeof(head)), edgenum  = 0;
		memset(du,0,sizeof(du));
		m=n-1;
		while(m--){
			int u,v;scanf("%d %d",&u,&v);
			addedge(u,v);
			addedge(v,u);
			du[u]++,du[v]++;
		}
		memset(vis,0,sizeof(vis));
		memset(use,0,sizeof(use));
		for(int i=1;i<=n;i++)dp[i] = 1;
		ans = 0;
		for(int i=1;i<=n;i++)
			if(du[i] == 1)
			{ vis[i]=1, dfs(i); break;}		
		
			if(ans>= n/K)printf("YES
");
			else printf("NO
");

	}
	return 0;
}
/*
2
4 2
1 2
2 3
3 4
4 2
1 2
1 3
1 4

*/