“旅行推销商问题”简易暴力实现

       应付离散实验足够了，但是还不会调用EasyX绘图啊~~~

#include <bits/stdc++.h>
#include <windows.h>

int start = 0;

// 城市名称
char city_name[6] = { 'A', 'B', 'C', 'D', 'E', 'F' };

// 记录到过的城市
int city[6] = {0};


int pass_count = 0;                // 记录当前经到过的城市数量
char pass_path[7] = {0};        // 记录路线


// 城市之间的路径
int city_path[6][6] =
{
    { 0, 6, 7, 2, 9, 16 },        // a
    { 6, 0, 3, 11, 12, 15 },    // b
    { 7, 3, 0, 9, 18, 5 },        // c
    { 2, 11, 9, 0, 13, 18 },    // d
    { 9, 12, 18, 13, 0, 13 },    // e
    { 16, 15, 5, 18, 13, 0 }    // f
};


// 判断是否已经到过a城市
bool ispassed( int a )
{
    return city[a] > 0;
}

void init()
{
    start = 0;
    pass_count = 0;
    int i;
    for( i = 0; i < 6; i ++ )
    {
        city[i] = 0;
        pass_path[i] = 0;
    }
    pass_path[i] = 0;
}

//////////////////////////////////////////////////////////////
// 功能：找出与当前城市最近且没有到达过的城市
// 参数：输入当前城市a,输出下一个城市b，
// 返回：两个城市之间的路程
//////////////////////////////////////////////////////////////
int calc_path( int a,int& b )
{
    int i =0;
    int min = 1000;
    b = 0;
    for(  i= 0; i < 6; i ++ )
    {
        if( min > city_path[a][i] && !ispassed(i) )
        {
            min = city_path[a][i];
            b = i;
        }
    }
    if( min < 1000 )
        city[b] = 1;
    return min;
}


bool input_city_num( int &city_num )
{
    char city_name = 'A';
    printf( "输入起点城市编号(A~F或a~f):" );
    scanf( "%c", &city_name );
    if( city_name >= 'a' && city_name <= 'f' )
        city_name -= 32;
    fflush(stdin);

    city_num = city_name - 'A' + 1;
    if( city_num > 7 || city_num <= 0 )
    {
        return false;
    }
    return true;
}


void start_travel()
{
    int city_num = 0;
    int i = 0;
    int path = 0;
    int ret = 0;
    int b;

    // 初始化数据
    init();

    // 输入起始城市编号
    while( !input_city_num(city_num) )
    {
        printf( "输入有误，请重新输入！
" );
    }

    city_num --;
    city[city_num] = 1;
    start = city_num;
    pass_path[ pass_count ++ ] = city_name[city_num];

    // 开始旅行规划
    printf( "
从城市[%c]出发
" , city_name[city_num] );
    printf( "
" );
    for( i = 0;i < 6; i ++ )
    {
        b = 0;
        ret = calc_path( city_num, b );
        if( ret > 0 && ret < 1000 )
        {
            path += ret;
            printf( "当前到达%c城市
", city_name[b] );
            printf( "当前走过路程：%d", path );
            printf( "
" );
            city_num = b;
            pass_path[ pass_count ++ ] = city_name[city_num];
        }
    }

    // 回到起点
    path += city_path[start][city_num];
    printf( "当前到达 %c 城市
", city_name[start] );
    printf( "当前走过路程：%d", path );
    printf("
");
    pass_path[pass_count++] = city_name[start];

    printf( "
" );
    printf( "旅行结束，总路程为：%d
路线：", path );
    for( i = 0; i < 7; i ++ )
    {
        if( i !=6 )
            printf( "%c--->", pass_path[i] );
        else
            printf( "%c", pass_path[i] );
    }
}

int main()
{
    while(1)
    {
        start_travel();
        printf("
----------------------------------------------------

");
    }
    return 0;
}