2019.01.24【POJ2449】Remmarguts' Date（K短路）

OpenJudge传送门

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BB:

我彻底放弃POJ了，它今天被卡成这样了。。。

各位以后还是去OpenJudge上做题吧，题号题目都和POJ完全一样的。

这道题就是K短路板题啊，可以看一下我的K短路讲解：https://blog.youkuaiyun.com/zxyoi_dreamer/article/details/86632445

嗯。我没用A*，直接用的可持久化左偏树。

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代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
#define ll long long
#define re register
#define gc get_char
#define cs const

namespace IO{
	inline char get_char(){
		cs static int Rlen=1<<20|1;
		static char buf[Rlen],*p1,*p2;
		return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,Rlen,stdin),p1==p2)?EOF:*p1++;
	}
	
	inline int getint(){
		re char c;
		while(!isdigit(c=gc()));re int num=c^48;
		while(isdigit(c=gc()))num=(num+(num<<2)<<1)+(c^48);
		return num;
	}
}
using namespace IO;

cs int N=1003,M=100005,B=7,INF=0x3f3f3f3f;
int n,m,k,S,T;

struct Graph{
	int last[N],nxt[M],to[M],w[M],ecnt;
	inline void addedge(int u,int v,int val){
		nxt[++ecnt]=last[u],last[u]=ecnt,to[ecnt]=v,w[ecnt]=val;
	}
}g,rg;

int dist[N];

inline void Dij(){
	memset(dist,0x3f,sizeof dist);
	set<pair<int,int> > q;
	q.insert(make_pair(dist[T]=0,T));
	cs int *last=rg.last,*nxt=rg.nxt,*w=rg.w,*to=rg.to;
	while(!q.empty()){
		re int u=q.begin()->second;q.erase(q.begin());
		for(int re e=last[u],v=to[e];e;v=to[e=nxt[e]]){
			if(dist[v]>dist[u]+w[e]){
				q.erase(make_pair(dist[v],v));
				dist[v]=dist[u]+w[e];
				q.insert(make_pair(dist[v],v));
			}
		}
	}
}

bool tree_edge[M],vis[N];
int fa[N],st[N],top;

void dfs(int u){
	vis[u]=true;
	st[++top]=u;
	for(int re e=rg.last[u],v=rg.to[e];e;v=rg.to[e=rg.nxt[e]])
	if(!vis[v]&&dist[v]==dist[u]+rg.w[e]){
		fa[v]=u;tree_edge[e]=true;
		dfs(v);
	}
}

namespace LT{
	int son[M*B][2];
	int ht[M*B],val[M*B],id[M*B];
	int tot;
	
	inline int newnode(int _val,int _id,int _dis=0){
		re int now=++tot;
		val[now]=_val,id[now]=_id;
		ht[now]=_dis,son[now][0]=son[now][1]=0;
		return now;
	}
	
	inline int _copy(int ori){
		re int now=++tot;
		val[now]=val[ori],id[now]=id[ori];
		ht[now]=ht[ori],son[now][0]=son[ori][0],son[now][1]=son[ori][1];
		return now;
	}
	
	inline int merge(int a,int b){
		if(!a||!b)return a|b;
		if(val[a]>val[b])swap(a,b);
		int now=_copy(a);
		son[now][1]=merge(son[now][1],b);
		if(ht[son[now][0]]<ht[son[now][1]])swap(son[now][0],son[now][1]);
		ht[now]=ht[son[now][1]]+1;
		return now;
	}
	
	inline void insert(int &rt,int val,int id){
		rt=merge(newnode(val,id),rt);
	}
}

int rt[M];

inline void build_heap(){
	int *last=g.last,*nxt=g.nxt,*to=g.to,*w=g.w;
	for(int re i=1;i<=top;++i){
		re int u=st[i];
		rt[u]=rt[fa[u]];
		for(int re e=last[u],v=to[e];e;v=to[e=nxt[e]])
		if(!tree_edge[e]&&dist[v]!=INF)LT::insert(rt[u],dist[v]-dist[u]+w[e],v);
	}
}

inline int solve_K(){
	if(!rt[S])return -1;
	priority_queue<pair<int,int>,vector<pair<int,int> > ,greater<pair<int,int> > > q;
	q.push(make_pair(dist[S]+LT::val[rt[S]],rt[S]));
	while(!q.empty()){
		pair<int,int> qq=q.top();q.pop();
		if((--k)==1)return qq.first;
		int v=qq.first,u=qq.second;
		int lc=LT::son[u][0],rc=LT::son[u][1],o=LT::id[u];
		if(rt[o])q.push(make_pair(v+LT::val[rt[o]],rt[o]));
		if(lc)q.push(make_pair(v+LT::val[lc]-LT::val[u],lc));
		if(rc)q.push(make_pair(v+LT::val[rc]-LT::val[u],rc));
	}
	return -1;
}

signed main(){
	n=getint();m=getint();
	for(int re i=1;i<=m;++i){
		int u=getint(),v=getint(),val=getint();
		g.addedge(u,v,val);
		rg.addedge(v,u,val);
	}
	S=getint(),T=getint(),k=getint();
	if(S==T)++k;
	Dij();
	if(dist[S]==INF){
		cout<<"-1";
		return 0;
	}
	if(k==1){
		cout<<dist[S];
		return 0;
	}
	dfs(T);
	build_heap();
	cout<<solve_K();
	return 0;
}