A - Til the Cows Come Home

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

90

Hint

INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

Bessie can get home by following trails 4, 3, 2, and 1.

Sponsor

提交poj疯狂给我ce报错，试了网上多个题解版本也一样，我直接默认我做对了吧 主要是怕之后找这个码找不到了 当成发个blog保存一下

/*Where there is light, in my heart.*/
/*SUMMER_TRAINING DAY 25*/
#include 
#include 
#include 
#include 
#include 
using namespace std;
//
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//#define INF 0x3f3f3f
#define ll long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define unmap(a,b) unordered_map
#define unset(a) unordered_set
#define F first
#define S second
#define pb push_back
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define _rep(i, a, b) for (int i = (a); i >= (b); --i)
#define mode 1e4+7
#define pi acos(-1)
#define U_queue priority_queue,greater>
typedef double db;
typedef pair<int,int> PII;
typedef pair PLL;
typedef vector<int> vi;
const int N = 2005;
//
int n,m;
int q[N],e[N],ne[N],dist[N],idx,h[N],w[N];
bool st[N];
//
void add(int a,int b,int c){
    e[idx]=b,w[idx]=c,ne[idx]=h[a],h[a]=idx++;
}
//
int dijkstra(){
    mem(dist,0x3f);
    U_queue q;
    q.push({0,1});
    dist[1]=0;
    dist[1]=0;
    while(q.size()){
        auto t=q.top();
        q.pop();
        int ver=t.S,d=t.F;
        if(st[ver]) continue;
        st[ver]=true;
        for(int i=h[ver];i!=-1;i=ne[i]){
            int j=e[i];
            if(dist[j]>dist[ver]+w[i])
            dist[j]=dist[ver]+w[i];
            q.push({dist[j],j});
        }
    }
    return dist[n];
}
signed main(){
    cin>>m>>n;
    mem(h,-1);
    while(m--){
        int a,b,c;
        cin>>a>>b>>c;
        add(a,b,c);
    }
    cout<<dijkstra()<
}
//made by shun 20220728