考研编程练习----最短路径问题

题目描述：

给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。

输入：

输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点t。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)

输出：

输出 一行有两个数， 最短距离及其花费。

样例输入：

3 2
1 2 5 6
2 3 4 5
1 3
0 0

样例输出：

　　　9 11

经典代码:

#include <stdio.h>

#define INT_MAX 0x7FFFFFFF

#define MAX_V 1001

#define MAX_E 1000

int main()

{

    int n, m, d[MAX_V], p[MAX_V], visited[MAX_V], path[MAX_E][4], s, t, minPos;

    while (scanf("%d%d", &n, &m) != EOF)

    {

        if (!n&&!m) break;

        //initialize

        for (int i = 0; i <= n; ++i)

        {

            d[i] = INT_MAX;

            p[i] = INT_MAX;

            visited[i] = 0;

        }

        //0 a,1 b,2 d,3 p

        for (int i = 0; i < m; ++i) scanf("%d%d%d%d", &path[i][0], &path[i][1], &path[i][2], &path[i][3]);

        scanf("%d%d", &s, &t);

        //algorithm start

        d[s] = 0; p[s] = 0; visited[s] = 1;

        for (int i = 0; i < m; ++i)

        {

            if (path[i][0] == s)

            {

                d[path[i][1]] = path[i][2];

                p[path[i][1]] = path[i][3];

            }

            else if (path[i][1] == s)

            {

                d[path[i][0]] = path[i][2];

                p[path[i][0]] = path[i][3];

            }

        }

        do

        {

            minPos = 0;

            for (int i = 1; i <= n; ++i)

            {

                if (visited[i]) continue;

                if (!minPos || d[i] < d[minPos] || (d[i] == d[minPos] && p[i] < p[minPos])) minPos = i;

            }

            if (minPos)

            {

                int a=minPos, b;

                for (int i = 0; i < m; i++)

                {

                    if (path[i][0] != a&&path[i][1] != a) continue;

                    if (path[i][0] == a) b = path[i][1];

                    else if (path[i][1] == a) b = path[i][0];

                    if (visited[b]) continue;

                    if (d[a] + path[i][2] < d[b]

                        || d[a] + path[i][2] == d[b] && p[a]+path[i][3]<p[b])

                    {

                        d[b] = d[a] + path[i][2];

                        p[b] = p[a] + path[i][3];

                    }

                }

                visited[a] = 1;

            }

        } while (minPos);

        printf("%d %d\n", d[t], p[t]);

    }

    return 0;

}