【BZOJ5506】[GXOI/GZOI2019]旅行者(最短路)

【BZOJ5506】[GXOI/GZOI2019]旅行者(最短路)

题面

BZOJ
洛谷

题解

正着做一遍\(dij\)求出最短路径以及从谁转移过来的,反过来做一遍,如果两个点不由同一个点转移过来就更新答案。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define ll long long
#define MAX 100100
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
struct Line{int v,next,w;}e[MAX*10];
int h[MAX],cnt=1;
inline void Add(int u,int v,int w){e[cnt]=(Line){v,h[u],w};h[u]=cnt++;}
int col[MAX];ll dis1[MAX],dis2[MAX];bool vis[MAX];
struct Node{int x,fr;ll dis;};
bool operator<(Node a,Node b){return a.dis>b.dis;}
priority_queue<Node>Q;
int n,m,K,a[MAX];ll ans;
void Dijkstra1()
{
	memset(dis1,63,sizeof(dis1));memset(vis,0,sizeof(vis));
	for(int i=1;i<=K;++i)dis1[a[i]]=0,col[a[i]]=a[i],Q.push((Node){a[i],a[i],0});
	while(!Q.empty())
	{
		Node p=Q.top();Q.pop();int u=p.x;
		if(vis[u])continue;vis[u]=true;
		for(int i=h[u];i;i=e[i].next)
		{
			int v=e[i].v;if(!(i&1))continue;
			if(dis1[v]>dis1[u]+e[i].w)
				col[v]=col[u],dis1[v]=dis1[u]+e[i].w,Q.push((Node){v,col[u],dis1[v]});
		}
	}
}
void Dijkstra2()
{
	memset(dis2,63,sizeof(dis2));memset(vis,0,sizeof(vis));
	for(int i=1;i<=K;++i)dis2[a[i]]=0,Q.push((Node){a[i],a[i],0});
	while(!Q.empty())
	{
		Node p=Q.top();Q.pop();int u=p.x;
		if(vis[u])continue;vis[u]=true;
		for(int i=h[u];i;i=e[i].next)
		{
			int v=e[i].v;if(i&1)continue;
			if(col[v]!=col[u])ans=min(ans,dis2[u]+e[i].w+dis1[v]);
			if(dis2[v]>dis2[u]+e[i].w)
				dis2[v]=dis2[u]+e[i].w,Q.push((Node){v,col[u],dis2[v]});
		}
	}
}
int main()
{
	int T=read();
	while(T--)
	{
		n=read();m=read();K=read();
		for(int i=1,u,v,w;i<=m;++i)	u=read(),v=read(),w=read(),Add(u,v,w),Add(v,u,w);
		for(int i=1;i<=K;++i)a[i]=read();
		ans=1e18;Dijkstra1();Dijkstra2();
		printf("%lld\n",ans);
		for(int i=1;i<=n;++i)h[i]=col[i]=0;cnt=1;
	}
	return 0;
}
posted @ 2019-04-19 15:17  小蒟蒻yyb  阅读(392)  评论(0编辑  收藏  举报