题意翻译
题目大意
给出一张图，请输出其中任意一条可行的从点 11 到点 nn 的最短路径。
输入输出格式
输入格式
第一行：两个整数n，m，分别表示点数和边数
接下来m行：每行三个整数u，v，w，表示u和v之间连一条边权为w的双向边。
输出格式
一行：一个可行的路径，如果不存在这种路径输出-1
2<=n<=10^5，0<=m<=10^5
题目描述
You are given a weighted undirected graph. The vertices are enumerated from 1 to nn . Your task is to find the shortest path between the vertex 11 and the vertex nn .
输入输出格式
输入格式：
The first line contains two integers nn and mm ( 2<=n<=10^{5},0<=m<=10^{5}2<=n<=105,0<=m<=105 ), where nn is the number of vertices and mm is the number of edges. Following mm lines contain one edge each in form a_{i}ai , b_{i}bi and w_{i}wi ( 1<=a_{i},b_{i}<=n,1<=w_{i}<=10^{6}1<=ai,bi<=n,1<=wi<=106 ), where a_{i},b_{i}ai,bi are edge endpoints and w_{i}wi is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
输出格式：
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
输入输出样例
输入样例#1： 复制
5 6
1 2 2
2 5 5
2 3 4
1 4 1
4 3 3
3 5 1
输出样例#1： 复制
1 4 3 5
输入样例#2： 复制
5 6
1 2 2
2 5 5
2 3 4
1 4 1
4 3 3
3 5 1
输出样例#2： 复制
1 4 3 5
分析：模板提
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=200005;
int n,m,st,en;
bool vis[N];//标记数组
ll dist[N];//距离数组
int pre[N];//前驱
int head[N],tot;
//结构体数组edge存边，edge[i]表示第i条边,
//head[i]存以i为起点的第一条边(在edge中的下标)
struct node
{
int next; //下一条边的存储下标
int to; //这条边的终点
ll w;//权值
} edge[N*4];
///tot为边的计数，从0开始计，每次新加的边作为第一条边，最后倒序遍历
void add(int u,int v,ll w)
{
edge[tot].to=v;
edge[tot].w=w;
edge[tot].next=head[u];
head[u]=tot++;
}
void init()
{
tot=0;
memset(head,-1,sizeof(head));
memset(pre,-1,sizeof(pre));
}
struct Rec
{
int id; //节点编号
ll dis; //距离
/*bool operator<(const Rec &tmp)const{
return dis>tmp.dis;
} */
};
bool operator <(const Rec&a,const Rec&b)
{
return a.dis>b.dis;
}
priority_queue<Rec> q;//重载小于号实现小根堆
void dijkstra()
{
for(int i=0; i<=n; i++)
dist[i]=1e18;
memset(vis,0,sizeof(vis));
dist[st]=0;
Rec rec;rec.id=st;rec.dis=0;
q.push(rec);
while(!q.empty())
{
int u=q.top().id;
q.pop();
if(vis[u])
continue;
vis[u]=1;
for(int i=head[u]; i!=-1; i=edge[i].next) //遍历以u为起点的所有边,与输入顺序相反
{
int v=edge[i].to;
ll w=edge[i].w;
if(dist[v]>dist[u]+w)
{
dist[v]=dist[u]+w;
Rec rec;
rec.id=v;
rec.dis=dist[v];
q.push(rec);
pre[v]=u;
}
}
}
}
vector<int> path;
void getPath()
{
for(int i=en; i!=-1; i=pre[i])
{
path.push_back(i);
}
}
int main()
{
scanf("%d%d",&n,&m);
st=1;en=n;
int u,v;
ll w;
init();
for(int i=1; i<=m; i++)
{
scanf("%d%d%lld",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
dijkstra();
if(dist[en]==1e18)
{
cout<<"-1"<<endl;
}
else
{
//cout<<dist[en]<<endl;
getPath();
for(int i=path.size()-1;i>=0;i--)
{
cout<<path[i]<<" ";
}
cout<<endl;
}
return 0;
}
