[Vjudge]一个人的旅行

题目描述如图
单源最短路裸题，只需要在所有节点之外建一个起点，让它指向草儿家附近的城市且路长为0，然后以这个点为起点跑一边最短路就可以了
代码

//By AcerMo
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lli long long int
using namespace std;
const int M=1050;
int t,s,d;
int dis[M],vis[M];
int to[M*3],nxt[M*3],head[M*3],val[M*3],cnt;
inline void read(int &x)
{
	x=0;char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) x=x*10+c-'0',c=getchar();
	return ;
}
inline void in_it()
{
	cnt=0;
	memset(dis,0x7f,sizeof(dis));
	memset(head,0,sizeof(head));
	memset(nxt,0,sizeof(nxt));
	return ;
}
inline void add(int x,int y,int w)
{
	to[++cnt]=y;nxt[cnt]=head[x];val[cnt]=w;head[x]=cnt;
	to[++cnt]=x;nxt[cnt]=head[y];val[cnt]=w;head[y]=cnt;
	return ;
}
inline void SPFA(int s)
{
	queue<int>q;q.push(s);
	vis[s]=1;dis[s]=0;
	while (q.size())
	{
		int x=q.front();
		q.pop();vis[x]=0;
		for (int i=head[x];i;i=nxt[i])
		{
			if (dis[to[i]]>dis[x]+val[i])
			{
				dis[to[i]]=dis[x]+val[i];
				if (vis[to[i]]==0) q.push(to[i]);
				vis[to[i]]=1;
			}
		}
	}
	return ;
}
signed main()
{
   	while (scanf("%d %d %d",&t,&s,&d)!=EOF)
   	{
   		in_it();
   		for (int i=1;i<=t;i++)
   		{
   			int x;read(x);
   			int y;read(y);
   			int w;read(w);
   			add(x,y,w);
		}
		for (int i=1;i<=s;i++)
		{
			int x;read(x);
			add(1001,x,0);
		}
		SPFA(1001);int ans=2147483647;
		for (int i=1;i<=d;i++)
		{
			int x;read(x);
			ans=min(ans,dis[x]);
		}
		cout<<ans<<endl;
		
	}
	return 0;
}

标签：旅行,cnt,一个,Vjudge,head,int,read,include,dis
来源： https://blog.csdn.net/ACerAndAKer/article/details/121261161