Dijkstra friends

Direct the examples:

HDU2544

In the annual school competition, all students finalists will receive a very nice t-shirt. But whenever our staff to hundreds of pieces of clothing shipped back to the stadium from the store when they are very tired! So now they want to find the shortest route to the stadium from the store, you can help them?

Input includes a plurality of sets of input data. The first line of each two integers N, M (N <= 100 , M <= 10000), N represents a few street intersections Chengdu, reference numeral 1 is to store the location of the intersection, the intersection designated N is the location of the stadium, M indicates a couple of choices in Chengdu. N = M = 0 indicates the end of input. Next M rows, each row comprising three integers A, B, C (1 < = A, B <= N, 1 <= C <= 1000), indicates there is a path between the intersection A and the intersection B, we staff need to C-minute time traveled this road.
Input line to ensure the presence of at least one track to store.
Output For each input and output line, represents the track workers come from a store minimum time Sample Input

Sample Output

3 
2 
Dijkstra always forget! ! ! !

#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring> 
using namespace std;
const int N=1E4+7;
const int INF=1e5+7;
vector<int >ve[N];
int arr[N][N];
int dis[N];
int mark[N];
int n,m;

void djstrea(int x){
    memset(mark,0,sizeof(mark)); //标记数组 
    for( Int I = . 1 ; I <= n-; I ++ ) { 
        DIS [I] = ARR [X] [I]; 
    } 
    
    DIS [X] = 0 ; // to its distance 0 
    Mark [X] = . 1 ; 
    
    
    for ( int I = . 1 ; I <= n; I ++) { // n points loop n times 
        int ANS = INF; // always find the minimum point of dis array 
        int POS;
         for ( int I = . 1 ; I <= n-; I ++ ) {
             IF (Mark [I] == 0 && ANS> DIS [I]) 
            { 
                ANS=dis[i];
                pos=i;
            }
        }
        
        
        mark[pos]=1;
        for(int k=1;k<=n;k++){
            if(!mark[k])
                dis[k]=min(dis[k],ans+arr[pos][k]);
        }
    }
}


int main(){
    while(cin>>n>>m){//n个点，m条边
        int x,y,z;
        if(n==0&&m==0) BREAK ;
         // initializes the maximum value 
        for ( int I = . 1 ; I <= n-; I ++ ) {
             for ( int J = . 1 ; J <= n-; J ++ ) 
                ARR [I] [J] = INF; 
        } 
        // patterning assignment 
        for ( int I = . 1 ; I <= m; I ++ ) { 
            CIN >> >> X >> Y Z; 
            ARR [X] [Y] = Z; 
            ARR [Y] [X] = Z; 
        } 
        
        int XX = 1 ; // find the shortest distance to the point xx
         djstrea (xx); 
        the printf ( " % D \ n- " , DIS [n-]); 
    } 
    
    return  0 ; 
}

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