CF1187E Tree Painting 换根法

思路

典型的树形(DP)

然而并不能一次(dfs)求出答案

要用换根法

按照个人理解的话

换根法就是

在选起点做树形(DP)时

对于(ans[1])//以(1)为起点的答案值

我们可以 通过(O(1)) 算出(ans[sonof1])

感觉就是树形(DP)嘛

实现

对于这道题

我们先以(1)为根

容易写出

inline void dfs(int x,int fa)
{
	size[x] = 1;
	for(reg i = head[x];i;i = edge[i].next)
	{
		int v = edge[i].v;
		if(v == fa) continue;
		dfs(v,x);
		size[x] += size[v];
	}
	ans[1] += size[x];
}

转移的话

先来张图

对于(x)

(ans[y]=f[xson1]+f[xson2]+f[xson3]+size[x]+size[y]+f[yson1]+f[yson2]+f[yson3])

有(size[x]=n-size[y])

(ans[y]=ans[x]+n-size[y]-size[y])

(code)

开(longlong)

#include <stack>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define reg register int
#define isdigit(x) ('0' <= x&&x <= '9')
template<typename T>
inline T Read(T Type)
{
	T x = 0,f = 1;
	char a = getchar();
	while(!isdigit(a)) {if(a == '-') f = -1;a = getchar();}
	while(isdigit(a)) {x = (x << 1) + (x << 3) + (a ^ '0');a = getchar();}
	return x * f;
}
const int MAXN = 2e5 + 10;
struct node
{
	int v,next;
}edge[MAXN << 1];
int cnt,size[MAXN],head[MAXN];
long long pr,ans[MAXN];
inline void addedge(int u,int v)
{
	edge[++cnt].v = v;
	edge[cnt].next = head[u];
	head[u] = cnt;
}
inline void dfs(int x,int fa)
{
	size[x] = 1;
	for(reg i = head[x];i;i = edge[i].next)
	{
		int v = edge[i].v;
		if(v == fa) continue;
		dfs(v,x);
		size[x] += size[v];
	}
	ans[1] += size[x];
}
inline void dfs2(int x,int fa)
{
	for(reg i = head[x];i;i = edge[i].next)
	{
		int v = edge[i].v;
		if(v == fa) continue;
		ans[v] = ans[x] + size[1] - 2 * size[v];
		pr = max(pr,ans[v]);
		dfs2(v,x);
	}
}
int main()
{
	int n = Read(1);
	for(reg i = 1;i < n;i++)
	{
		int u = Read(1),v = Read(1);
		addedge(u,v),addedge(v,u);
	}
	dfs(1,0);
	dfs2(1,0);
	printf("%lld",pr);
	return 0;
}

//f[x] = n + f[xson]

//f[y] = n + f[yson] = f[x] + n - 2 * size[y]