BZOJ.1576.[Usaco2009 Jan]安全路经Travel(树形DP 并查集)

题目链接 BZOJ
洛谷

先求最短路树。考虑每一条非树边(u,v,len),设w=LCA(u,v),这条边会对w->v上的点x(x!=w)有dis[u]+dis[v]-dis[x]+len的距离。
每条边用dis[u]+div[v]+len更新链。树剖就做完了。
因为每个点只需取最小值,所以把边按dis[u]+div[v]+len排序后并查集更新链也行。
复杂度(O(nalpha(n)+mlogm))

树DP失败,好像没法处理子树内的ndis互相更新。。唉。

//12680kb	556ms
#include <queue>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
//#define gc() getchar()
#define MAXIN 300000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
#define mp std::make_pair
#define pr std::pair<int,int>
const int N=1e5+5,M=4e5+5,INF=0x3f3f3f3f;

int n,fa[N],ff[N],Enum,H[N],nxt[M],to[M],len[M],dis[N],ans[N];
char IN[MAXIN],*SS=IN,*TT=IN;
struct Edge
{
	int u,v,w;
	bool operator <(const Edge &x)const{
		return w<x.w;
	}
}e[M];

inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now;
}
inline void AE(int w,int u,int v)//order
{
	to[++Enum]=v, nxt[Enum]=H[u], H[u]=Enum, len[Enum]=w;
	to[++Enum]=u, nxt[Enum]=H[v], H[v]=Enum, len[Enum]=w;
}
void Dijkstra()
{
	static bool vis[N];
	static std::priority_queue<pr> q;
	memset(dis,0x3f,sizeof dis);
	q.push(mp(dis[1]=0,1));
	while(!q.empty())
	{
		int x=q.top().second; q.pop();
		if(vis[x]) continue;
		vis[x]=1;
		for(int i=H[x],v; i; i=nxt[i])
			if(dis[v=to[i]]>dis[x]+len[i])
				fa[v]=x, q.push(mp(-(dis[v]=dis[x]+len[i]),v));
	}
}
inline int Find(int x)
{
	return x==ff[x]?x:ff[x]=Find(ff[x]);	
}
void Update(int u,int v,int val)
{
	u=Find(u), v=Find(v);
	while(u!=v)
	{
		if(dis[u]<dis[v]) std::swap(u,v);//没必要用dep,保证在一条链上时上下不颠倒就行了 
		ans[u]=val-dis[u];
		u=ff[u]=Find(fa[u]);
	}
}

int main()
{
	n=read(), Enum=1;
	for(int m=read(),i=1; i<=m; ++i) AE(read(),read(),read());
	Dijkstra();
	int t=0;
	for(int i=2,lim=Enum; i<=lim; i+=2)
	{
		int u=to[i], v=to[i^1];
		if(fa[u]==v||fa[v]==u) continue;
		e[++t]=(Edge){u,v,dis[u]+dis[v]+len[i]};
	}
	std::sort(e+1,e+1+t);
	for(int i=1; i<=n; ++i) ff[i]=i;
	for(int i=1; i<=t; ++i) Update(e[i].u,e[i].v,e[i].w);
	for(int i=2; i<=n; ++i) printf("%d
",ans[i]?ans[i]:-1);

	return 0;
}

纪念一下10分的树DP

#include <queue>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cstring>
#include <algorithm>
#define gc() getchar()
#define mp std::make_pair
#define pr std::pair<LL,int>
#define Vec std::vector<int>
typedef long long LL;
const int N=1e5+5,M=4e5+5;
const LL INF=0x3f3f3f3f3f3f3f3f;

int n,pre[N],dep[N],fa[N],sz[N],son[N],in[N],out[N],ref[N],Index,top[N];
LL dis[N],ndis[N],ans[N];
bool is_t[M];
Vec vec[N];
bool de;
struct Graph
{
	int Enum,H[N],nxt[M],fr[M],to[M];
	LL len[M];
	inline void AE(int w,int u,int v)//order
	{
		to[++Enum]=v, fr[Enum]=u, nxt[Enum]=H[u], H[u]=Enum, len[Enum]=w;
		to[++Enum]=u, fr[Enum]=v, nxt[Enum]=H[v], H[v]=Enum, len[Enum]=w;
	}
	inline void AE_d(int u,int v,LL w)
	{
		to[++Enum]=v, nxt[Enum]=H[u], H[u]=Enum, len[Enum]=w;
	}
}T,G;

inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now;
}
void Dijkstra()
{
	static bool vis[N];
	static std::priority_queue<pr> q;
	memset(dis,0x3f,sizeof dis);
	q.push(mp(dis[1]=0,1));
	while(!q.empty())
	{
		int x=q.top().second; q.pop();
		if(vis[x]) continue;
		vis[x]=1;
		for(int i=G.H[x],v; i; i=G.nxt[i])
			if(dis[v=G.to[i]]>dis[x]+G.len[i])
				pre[v]=i, q.push(mp(-(dis[v]=dis[x]+G.len[i]),v));
	}
	for(int i=2; i<=n; ++i) is_t[pre[i]]=1, T.AE_d(G.fr[pre[i]],i,G.len[pre[i]]);
}
void DFS1(int x)
{
	int mx=0; sz[x]=1;
	for(int i=T.H[x],v; i; i=T.nxt[i])
	{
		fa[v=T.to[i]]=x, dep[v]=dep[x]+1, DFS1(v), sz[x]+=sz[v];
		if(sz[v]>mx) mx=sz[v], son[x]=v;
	}
}
void DFS2(int x,int tp)
{
	ref[in[x]=++Index]=x, top[x]=tp;
	if(son[x])
	{
		DFS2(son[x],tp);
		for(int i=T.H[x]; i; i=T.nxt[i])
			if(T.to[i]!=son[x]) DFS2(T.to[i],T.to[i]);
	}
	out[x]=Index;
}
inline int LCA(int u,int v)
{
	while(top[u]!=top[v]) dep[top[u]]>dep[top[v]]?u=fa[top[u]]:v=fa[top[v]];
	return dep[u]>dep[v]?v:u;
}
inline int LCA2(int u,int lca)
{
	int las=u;
	while(top[u]!=top[lca]) u=fa[las=top[u]];
	return u==lca?las:ref[in[lca]+1];
}
LL Solve(int x)
{
	if(de) printf("
Begin!
ans[%d]=%I64d
",x,ans[x]);
	for(int i=G.H[x],v; i; i=G.nxt[i])
		if((v=G.to[i])!=fa[x]&&(in[v=G.to[i]]<in[x]||out[v]>out[x]))
			de&&(printf("ans:%d->%d:%I64d
",v,x,dis[v]+G.len[i])),
			ans[x]=std::min(ans[x],dis[v]+G.len[i]);
	if(de) printf("update by other tree:ans[%d]=%I64d
",x,ans[x]);
	for(int i=0,l=vec[x].size(); i<l; ++i)
	{
		int id=vec[x][i],u=G.fr[id],v=G.to[id];
		de&&(printf("ndis:%d->%d:%I64d
",u,v,dis[u]+G.len[id]));
		ndis[v]=std::min(ndis[v],dis[u]+G.len[id]);
	}
	LL mn=INF;
	for(int i=T.H[x],v; i; i=T.nxt[i])
		mn=std::min(mn,Solve(T.to[i])+T.len[i]);
	ndis[x]=std::min(ndis[x],mn), ans[x]=std::min(ans[x],ndis[x]);
	if(de) printf("End ans[%d]=%I64d ndis[%d]=%I64d mn=%I64d
",x,ans[x],x,ndis[x],mn);
	return ndis[x];
}

int main()
{
	freopen("C.in","r",stdin);
	freopen("C.out","w",stdout);

	n=read(), G.Enum=1;
	for(int m=read(),i=1; i<=m; ++i) G.AE(read(),read(),read());
	Dijkstra(), DFS1(1), DFS2(1,1);
	de=0;
	if(de) for(int i=1; i<=n; ++i) printf("%d:fa:%d in:%d out:%d top:%d
",i,fa[i],in[i],out[i],top[i]);
	for(int i=2,lim=G.Enum; i<=lim; ++i)
	{
		if(is_t[i]||is_t[i^1]) continue;
		int u=G.fr[i], v=G.to[i], w=LCA(u,v), w2=LCA2(v,w);
//		printf("LCA2(%d,%d):%d
",u,v,w2);
		vec[w2].push_back(i);
	}
	memset(ans,0x3f,sizeof ans);
	memset(ndis,0x3f,sizeof ndis);
	Solve(1);
	for(int i=2; i<=n; ++i) printf("%I64d
",ans[i]==INF?-1ll:ans[i]);

	return 0;
}
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无心插柳柳成荫才是美丽
有哪种美好会来自于刻意
这一生波澜壮阔或是不惊都没问题
只愿你能够拥抱那种美丽
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原文地址:https://www.cnblogs.com/SovietPower/p/9613561.html