POJ3585 Accumulation Degree 【树形dp】

题目链接

POJ3585

题解

-二次扫描与换根法-

对于这样一个无根树的树形dp

我们先任选一根进行一次树形dp

然后再扫一遍通过计算得出每个点为根时的答案

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#define LL long long int

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define cls(s) memset(s,0,sizeof(s))

using namespace std;

const int maxn = 200005,maxm = 400005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int h[maxn],ne = 2,de[maxn];

struct EDGE{int to,nxt,w;}ed[maxm];

inline void build(int u,int v,int w){

	ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;

	ed[ne] = (EDGE){u,h[v],w}; h[v] = ne++;

	de[u]++; de[v]++;

}

int n,fa[maxn];

int d[maxn],f[maxn],g[maxn],ans;

void dfs1(int u){

	f[u] = 0;

	if (de[u] == 1 && u != 1) return;

	Redge(u) if ((to = ed[k].to) != fa[u]){

		fa[to] = u; d[to] = ed[k].w; dfs1(to);

		if (de[to] == 1) f[u] += ed[k].w;

		else f[u] += min(ed[k].w,f[to]);

	}

}

void dfs2(int u){

	g[u] = f[u];

	if (fa[u]){

		if (de[fa[u]] == 1) g[u] += d[u];

		else g[u] += min(d[u],g[fa[u]] - min(d[u],f[u]));

	}

	ans = max(ans,g[u]);

	Redge(u) if ((to = ed[k].to) != fa[u]){

		dfs2(to);

	}

}

int main(){

	int T = read();

	while (T--){

		ne = 2; cls(h); cls(de); ans = 0;

		n = read(); int a,b,w;

		for (int i = 1; i < n; i++){

			a = read(); b = read(); w = read();

			build(a,b,w);

		}

		dfs1(1);

		//REP(i,n) printf("%lld ",f[i]); puts("");

		dfs2(1);

		//REP(i,n) printf("%lld ",g[i]); puts("");

		printf("%d\n",ans);

	}

	return 0;

}