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
",ans);
	}
	return 0;
}