Description

給定一棵樹求最小路徑覆蓋

Solution

我們需要用上這是一棵樹的性質。設f[x]表示x為根的子樹內的最小路徑覆蓋且x是一條路徑的端點，g[x]表示x為根的子樹內最小路徑覆蓋且x不是一條路徑的端點。

Code

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define
 
 fill(x,t) memset(x,t,sizeof(x))

const int INF=0x3f3f3f3f;
const int N=20005;
const int E=500005;

struct edge {int y,next;} e[E];

int ls[N],edCnt=1;
int f[N],g[N];

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
	for (;ch<='9'&&ch>=
 
'0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

void add_edge(int x,int y) {
	e[++edCnt]=(edge) {y,ls[x]}; ls[x]=edCnt;
	e[++edCnt]=(edge) {x,ls[y]}; ls[y]=edCnt;
}

void dfs(int now,int fa) {
	f[now]=g[now]=INF;
	int rec=INF,sum=0,cnt=0;
	for (int i=ls[now];i;i=e[i].next) {
		if (e[i].y==fa) continue;
		dfs
 
(e[i].y,now); cnt++;
		sum+=std:: min(f[e[i].y],g[e[i].y]);
		rec=std:: min(rec,std:: max(0,f[e[i].y]-g[e[i].y]));
	}
	f[now]=sum+std:: min(rec,1);
	if (cnt<2) return ;
	int mn1=INF,mn2=INF;
	for (int i=ls[now];i;i=e[i].next) {
		if (e[i].y==fa) continue;
		int tmp=std:: max(0,f[e[i].y]-g[e[i].y]);
		if (tmp<mn1) mn2=mn1,mn1=tmp;
		else if (tmp<mn2) mn2=tmp;
	}
	g[now]=sum+mn1+mn2-1;
}

int main(void) {
	freopen("data.in","r",stdin);
	freopen("myp.out","w",stdout);
	for (int T=read();T--;) {
		edCnt=1; fill(ls,0);
		int n=read();
		rep(i,2,n) add_edge(read(),read());
		dfs(1,0);
		printf("%d\n", std:: min(f[1],g[1]));
	}
	return 0;
}