dijkstra算法--寻找最短路径

#include<iostream>
using namespace std;
#define INF 100
//0:北区宿舍
//1：第二食堂
//2：2区
//3：3区
//4：4区
//5：5区
//6：6区
//7：7区
//8：8区
//9：图书馆

//图的邻接矩阵
int matrix[10][10] = {
    0,1,3,23,INF,15,6,7,28,12,
    0,0,12,2,13,3,14,6,INF,10,
    0,0,0,11,1,21,INF,24,5,38,
    0,0,0,0,11,13,4,25,6,38,
    0,0,0,0,0,22,12,14,INF,18,
    0,0,0,0,0,0,INF,12,4,48,
    0,0,0,0,0,0,0,22,24,18,
    0,0,0,0,0,0,0,0,21,26,
    0,0,0,0,0,0,0,0,0,26,
    0,0,0,0,0,0,0,0,0,0
};
//各个景点名字
char name[10][100] = {
    "北区宿舍",
    "第二食堂",
    "二区教学楼",
    "三区教学楼",
    "四区教学楼",
    "五区教学楼",
    "六区教学楼",
    "七区教学楼",
    "八区教学楼",
    "图书馆"
};
//景点的详细内容
char content[10][500] = {
    "提供住宿",
    "提供饮食",
    "教学区2",
    "教学区3",
    "教学区4",
    "教学区5",
    "教学区6",
    "教学区7",
    "教学区8",
    "提供阅读，书籍借阅与归还"
};

//S的作用是记录已求出最短路径的顶点(以及相应的最短路径长度)，
//而U则是记录还未求出最短路径的顶点(以及该顶点到起点s的距离)。
int s[10], u[10];


bool flag[10] = { false };//用以标志已经存储在s[]中的点


//找到与i节点相连路径的最短节点   返回其下标
int min(int i) {
    int min = INF;
    int tag = 0;
    for (int j = 0; j < 10; j++)
    {
        if (j == i)  //不能返回自身  即权重为0的点
            continue;
        else if(!flag[j])  //如果没被标志
        {
            if (min > matrix[i][j])
            {
                min = matrix[i][j];
                tag = j;
            }
                
        }
    }
    return tag;
}
//介绍图景
void serveintroduce()
{
    int i;
    cout <<endl<< "please input your choice to get more information:";
    cin >> i;
    while (i > 9 || i < 0)
    {
        cout << "please input the true number!!!";
        cout << "please input your choice to get more information:";
        cin >> i;
    }

    cout << "the number you have input is  :" << i << endl;
    cout << "the building of the number is :" << name[i] << endl;
    cout << "this building‘s function is   :" << content[i] << endl;
}
//找到最短路径  输入出发点与目的地  输出最短路径长度
void findtrade()
{
    int st, en;   //开始节点st   结束节点en
    int pre[10];  //用以记录起始点到各点最短路径的前驱
    memset(s, 0, sizeof(s));
    memset(u, 0, sizeof(u));
    cout << endl << "please input the start point and destination:" << endl;
    cin >> st;
    cin >> en;
    while (st > 9 || st < 0 || en>9 || en < 0)
    {
        cout << "please input the true number!!!";
        cout << endl << "please input the start point and destination:" << endl;
        cin >> st;
        cin >> en;
    }
    if (st == en)
    {
        cout << "原地踏步";
        return;
    }

    for (int i = 0; i < 10; i++)
    {
        s[i] = u[i] = matrix[st][i];
        pre[i] = st;         //前驱
        flag[i] = false;     //刷新标志位  每次执行findtrade（）函数都需要刷新标志
    }
    

    //Dijkstra算法
    int m = st;
    flag[st] = true;
    for (int i = 0; i < 10; i++)
    {
        if (m == en)
            break;
        if (s[min(m)] > s[m] + matrix[m][min(m)])
        {
            pre[min(m)] = m;
            s[min(m)] = s[m] + matrix[m][min(m)];
        }
        m = min(m);
        flag[m] = true;
    }
    cout << name[st] << "到" << name[en] << "的最短距离为" << s[en] << endl;   //输出最短距离

    m = en;
    cout << name[en];
    while (pre[m] != st)   //打印最短路径
    {
        cout << "<---" << name[pre[m]];
        m = pre[m];
    }
    cout << "<---" << name[st] << endl;
}

int main()
{
    int i = 0;
    cout << "plaese read the review to choose what you want to know" << endl;
    for (i = 0; i < 10; i++)
    {
        cout << i << " :";
        cout << name[i] << " " << endl;
    }

    for (int i = 0; i < 10; i++)
    {
        for (int j = 9; j > i; j--)
            matrix[j][i] = matrix[i][j];
    }

    //for(int i = 0; i < 10; i++)
    //{
    //    for (int j = 0; j < 10; j++)
    //        cout << matrix[i][j]<<" ";
    //    cout << endl;
    //}

    while (i)
    {
        cout << endl<<"please input the num to choose the following choice:" << endl;
        cout << "1, service of introduce; 2, find the trade" << endl;
        cin >> i;
        while (i != 1 && i != 2)
        {
            cout << endl<<"please input the true number!!!" << endl;
            cout << "please input the num to choose the following choice:" << endl;
            cout << "1, service of introduce; 2, find the trade" << endl;
            cin >> i;
        }
        if (i == 1)
            serveintroduce();
        else
            findtrade();
    }

    return 0;
}