Detailed explanation of the implementation of Call, Apply, and Bind in JS (source code implementation)

2023-01-25   ES  

Delete the connection between point A and B, and then find the shortest circuit between the two points. If there is the shortest circuit, and the length between A and B is the minimum value, which is the shortest ring required. But I also asked for the output path. I used SPFA to find the shortest circuit and record the path.

After finishing this method, I saw that many of the short rings were found in Discuss, so I searched the Internet. It turned out that Flowd could directly find the shortest ring in the time of O (N^3), but this labeling path was a bit troublesome. In the end, he did not think of a better way to record the path, but he did find the shortest ring.

SPFA Seeking the shortest road algorithm:



View Code

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <math.h>
#include <map>
#include <stack>
#include <vector>
#define  N 104
#define  INF 1000000000
using namespace std ;

int mp[N][N] , pre[N] , step[N] ;
int dis[N] , n , m , ans ;
bool f[N] ;
queue<int>q ;

void init()
{
    memset( f , false , sizeof( f )) ;
    memset( pre , -1 , sizeof ( pre )) ;
    for ( int i = 0 ; i <= n ; i++ )
    dis[i] = INF ;
    while( !q.empty()) q.pop();
}

int Spfa( int s , int e )
{
    init() ;
    f[s] = true ;
    dis[s] = 0 ;
    q.push( s ) ;

    while ( !q.empty())
    {
        int u = q.front() ;
        q.pop();
        f[u] = false ;

        for ( int i = 1 ; i <= n ; i++ )
        {
            if ( mp[u][i] < INF && dis[i] > dis[u] + mp[u][i] )
            {
                dis[i] = dis[u] + mp[u][i] ;
                pre[i] = u ;

                if ( !f[i] )
                {
                    f[i] = true ;
                    q.push( i ) ;
                }
            }
        }
    }
    return dis[e] ;
}

void output( int x )
{
    if ( step[x] == -1 )
    return ;

    output( step[x] ) ;
    printf ( "%d " , step[x] ) ;
    return ;
}

int main()
{
    int i , j , x , y , z ;

    while ( scanf ( "%d" , &n ) , n != -1 )
    {
        for ( i = 0 ; i <= n ; i++ )
        {
            for ( j = 0 ; j <= n ; j++ )
            mp[i][j] = INF ;
        }
        scanf ( "%d" , &m ) ;
        for( i = 1 ; i <= m ; i++ )
        {
            scanf ( "%d%d%d" , &x , &y , &z ) ;
            if ( mp[x][y] > z )
            mp[x][y] = mp[y][x] = z ;
        }

        memset( step , -1 , sizeof ( step )) ;
        int pos = -1 ;
        ans = INF ;
        for ( i = 1 ; i <= n ; i++ )
        {
            for ( j = 1 ; j <= n ; j++ )
            {
                if ( mp[i][j] < INF )
                {
                    z = mp[i][j] ;
                    mp[i][j] = mp[j][i] = INF ;
                    y = Spfa( j , i ) + z ;
                    if ( y < ans )
                    {
                        ans = y ;
                        pos = i ;
                        for ( x = 1 ; x <= n ; x++ )
                        step[x] = pre[x] ;
                    }
                }
            }
        }
        if( pos == -1 )
        {
            printf ( "No solution.\n" ) ;
        }
        else
        {
            output( pos ) ;
            printf ( "%d\n" , pos ) ;
        }
    }
    return 0 ;
}

has been wa several times at the first sample. I do n’t know why, I do n’t know if it is related to the order of path output, but I change the order output of the path. Essence Essence Essence Essence Essence Essence

FLOYD seek the shortest ring (no output path)



View Code

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <math.h>
#include <map>
#include <stack>
#include <vector>
#define  N 104
#define  INF 1000000000
using namespace std ;

int dis[N][N] , grap[N][N] ;
int n , m , ans ;
int path[N][N] ;

void output( int s , int e )
{
    if ( s == e )
    return ;
    if ( path[s][e] == -1 )
    {
        printf ( " %d" , e ) ;
        return ;
    }
    else
    {
        output( s , path[s][e] ) ;
        output( path[s][e] , e ) ;
    }
}

int main()
{
    int i , j , k , x , y , z ;

    while ( scanf ( "%d" , &n ) , n != -1 )
    {
        for ( i = 1 ; i <= n ; i++ )
        {
            for ( j = 1 ; j <= n ; j++ )
            {
                dis[i][j] = grap[i][j] = INF ;
                path[i][j] = -1 ;
            }
        }

        scanf ( "%d" , &m ) ;
        for ( i = 1 ; i <= m ; i++ )
        {
            scanf ( "%d%d%d" , &x , &y , &z ) ;
            if ( grap[x][y] > z )
            {
                grap[x][y] = grap[y][x] = z ;
                dis[x][y] = dis[y][x] = z ;
            }
        }

        //int s , e ;
        ans = INF ;
        for ( k = 1 ; k <= n ; k++ )
        {
            for ( i = 1 ; i < k ; i++ )
            {
                for ( j = 1 ; j < i ; j++ )
                if( ans > dis[i][j] + grap[i][k] + grap[k][j] )
                {
                    ans = dis[i][j] + grap[i][k] + grap[k][j] ;
                    //path[i][j] = k ;
                    //s = j ; e = i ;
                }
            }

            for ( x = 1 ; x <= n ; x++ )
            {
                for ( y = 1 ; y < x ; y++ )
                if( dis[x][k] + dis[k][y] < dis[x][y] )
                {
                    dis[x][y] = dis[x][k] + dis[k][y] ;
                    //path[x][y] = k ;
                }
            }
        }

        if ( ans == INF )
        cout<<"No solution."<<endl ;
        else
        {
            /*i = 0 ;
            int step[N] ;
            for ( k = s ; k != e ; k = path[k][e] )
            {
                step[i++] = k ;
            }
            step[i++] = e ;
            step[i++] = k ;

            for ( j = 0 ; j < i ; j++ )
            cout<<step[j]<<" ";
            cout<<endl;*/
            cout<<ans<<endl ;
        }
    }
    return 0 ;
}

Reprinted: https://www.cnblogs.com/misty1/archive/2013/01/25/2877272.html

source

Related Posts

[bzoj2157] Travel tree chain section

Struts 2 traversed tags for Iterator usage

mount and umount commands (mounting and uninstalling file system) Deepwater

asp.net: Basic drawing + add watermark to the picture

Detailed explanation of the implementation of Call, Apply, and Bind in JS (source code implementation)

Random Posts

SUBLIME Text formatting JSON-PRETTY JSON

MongoDB installation configuration and PyMongo installation and use

Tools -simple Java realizes pure text emails (built using SpringBoot)

Insert data to the database, automatically generate time when modifying data, and automatic modification time

ROBOGUIDE software: Robot conveyor belt up and down feeding virtual simulation operation method