hdu 1874 continued unimpeded engineering (SPFA template)

Smooth Traffic Project Continued

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 82186    Accepted Submission(s): 31619

Problem Description

Since the province of smooth implementation of the project for many years, finally built many roads. But more is not a good way, each time from one town to another, there are many kinds of road schemes to choose from, and some programs are other programs, walking distance much shorter than others. This allows pedestrians very troubled.

Now, start and end points are known, you calculate from the beginning to the end, how much the shortest distance required to walk.

 

 

Input

This topic contains multiple sets of data, the processing to the end of the file.
Each test line contains two positive integers N and M (0 <N <200,0 < M <1000), represents the number of towns and the number of existing roads have been built. In urban numbered 0 ~ N-1.
Followed by M lines road information. Each row contains three integers A, B, X (0 < = A, B <N, A! = B, 0 <X <10000), expressed between towns and urban A B X a length of two-way road.
The next line followed by two integers S, T (0 <= S , T <N), representing the start and end points.

 

 

Output

For each set of data, the output from the minimum required to walk on one line. If the route from S to T does not exist, the output of -1.

 

 

Sample Input

3 3

0 1 1

0 2 3

1 2 1

0 2

3 1

0 1 1

1 2

 

 

Sample Output

2

-1

 

#include <the iostream> 
#include < String .h> 
#include < String > 
#include <algorithm> 
#include <Queue>
 #define LL Long Long
 #define MX 0x3f3f3f3f
 the using  namespace STD;
 int Way [ 250 ] [ 250 ], DIS [ 250 ], VIS [ 250 ], CNT [ 250 ];
 // Way recording path relationship, dis [i] records the closest distance from the point j, vis [i] whether the marker in the queue, cnt [i] records i into the number of points the queue 
int n-, m;
 void the init () 
{ 
    for ( int i =0 ; I <n-; I ++) // initialize Way 
    {
         for ( int J = 0 ; J <n-; J ++ ) 
        { 
            IF (I == J) 
                Way [I] [J] = 0 ;
             the else 
                Way [I] [J] = MX; 
        } 
    } 
} 

void SPFA ( int ST) 
{ 
    for ( int I = 0 ; I <n-; I ++) // this point number is from 0 to 
        DIS [I] = MX; 
    Memset (VIS, 0 , sizeof(vis));
    memset(cnt,0,sizeof(cnt));
    vis[st]=1;
    cnt[st]=1;
    dis[st]=0;
    queue<int>p;
    p.push(st);
    while(!p.empty())
    {
        int now=p.front();
        p.pop();
        vis[now]=0;
        for(int i=0;i<n;i++)
        {
            if(dis[now]+way[now][i]<DIS [I]) 
            { 
                DIS [I] = DIS [now] + Way [now] [I];
                 IF (VIS [I] == 0 ) // if the point is not in queue inside 
                { 
                    p.push (I); 
                    VIS [I] = . 1 ; 
                    CNT [I] ++ ;
                     IF (CNT [I]> n) // If this point to add more than n times, indicating a negative ring directly return 
                        return ; 
                } 
            } 
        } 
    } 

} 
int main () 
{ 
    int ST, End;
     the while(~ Scanf ( " % D% D " , & n-, & m)) // input to the end, or TLE 
    { 
        the init (); // initialize Way 
        for ( int I = 0 ; I <m; I ++ ) 
        { 
            int X, Y, Z; 
            Scanf ( " % D% D% D " , & X, & Y, & Z);
             IF (Way [X] [Y]> Z) // multiple edges 
            { 
                Way [X] [Y] = Z; 
                Way [Y] [X] = Z; 
            } 

        } 
        Scanf ( " % D% D" , & ST, & End); 
        SPFA (ST); // SPFA can be obtained starting point to each of the shortest distance 
        IF (DIS [End] == MX) 
            the printf ( " -1 \ n- " );
         the else 
            the printf ( " % D \ n- " , DIS [End]); 
    } 
    return  0 ; 
}