BZOJ4890 [Tjoi2017]城市 【树形dp】

题目链接

BZOJ4890

题解

枚举断开哪一条边，然后对剩余的两棵树分别做一遍换根法树形dp
需要求出每个点到树中其它点距离的最大值(f[i])和次大值(g[i])【用以辅助换根计算最大值】
求出每棵树中的最长路径，然后再将两棵树中(f[i])最小值相连保证相连后产生的最大值最小

(O(n^2))的复杂度
如果怕被卡常，可以知道要切的边一定在直径上，虽然上界没有变，但速度可以快不少

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#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 mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 5005,maxm = 100005,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 = 1;
struct EDGE{int from,to,nxt,w,flag;}ed[maxn << 1];
inline void build(int u,int v,int w){
	ed[++ne] = (EDGE){u,v,h[u],w,1}; h[u] = ne;
	ed[++ne] = (EDGE){v,u,h[v],w,1}; h[v] = ne;
}
int f[maxn],g[maxn],ch[maxn],d[maxn],du,path[maxn],fa[maxn];
int n,ans,fans,mnans;
void DFS(int u,int Fa){
	if (d[u] > d[du]) du = u;
	Redge(u) if ((to = ed[k].to) != Fa){
		d[to] = d[u] + ed[k].w; path[to] = k;
		DFS(to,u);
	}
}
void dfs(int u){
	f[u] = g[u] = 0;
	Redge(u) if (ed[k].flag && (to = ed[k].to) != fa[u]){
		fa[to] = u; d[to] = ed[k].w; dfs(to);
		if (f[to] + ed[k].w > f[u]){
			g[u] = f[u];
			f[u] = f[to] + ed[k].w;
			ch[u] = to;
		}
		else if (f[to] + ed[k].w > g[u]) g[u] = f[to] + ed[k].w;
	}
	fans = max(fans,f[u]);
}
void dfs2(int u){
	if (fa[u]){
		int v = fa[u];
		if (ch[v] == u){
			if (g[v] + d[u] > f[u]){
				g[u] = f[u];
				f[u] = g[v] + d[u];
				ch[u] = v;
			}
			else if (g[v] + d[u] > g[u]) g[u] = g[v] + d[u];
		}
		else {
			if (f[v] + d[u] > f[u]){
				g[u] = f[u];
				f[u] = f[v] + d[u];
				ch[u] = v;
			}
			else if (f[v] + d[u] > g[u]) g[u] = f[v] + d[u];
		}
	}
	Redge(u) if (ed[k].flag && (to = ed[k].to) != fa[u]){
		dfs2(to);
	}
	fans = max(fans,f[u]);
	mnans = min(mnans,f[u]);
}
int dp(int rt){
	d[rt] = 0; fa[rt] = 0; dfs(rt);
	mnans = INF;
	dfs2(rt);
	return mnans;
}
int main(){
	n = read(); int a,b,w; ans = INF;
	for (int i = 1; i < n; i++){
		a = read(); b = read(); w = read();
		build(a,b,w);
	}
	d[du = 1] = 0; DFS(1,0);
	path[du] = 0; d[du] = 0; DFS(du,0);
	int t1,t2;
	for (int i = du; path[i]; i = ed[path[i]].from){
		int k = path[i];
		ed[k].flag = ed[k ^ 1].flag = false;
		cls(f); cls(g); fans = 0;
		t1 = dp(ed[k].from);
		t2 = dp(ed[k].to);
		//REP(j,n) printf("%d ",f[j]); puts("");
		fans = max(fans,t1 + t2 + ed[k].w);
		//printf("(%d,%d)   mx = %d
",ed[k].to,ed[k].from,fans);
		ans = min(ans,fans);
		ed[k].flag = ed[k ^ 1].flag = true;
	}
	printf("%d
",ans);
	return 0;
}