hdu3790 - 最短路径问题 (Dijkstra)(多条最短路径找花费最少的一条)

时间:2022-02-14 09:41:21

最短路径问题

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13771    Accepted Submission(s): 4234


Problem Description
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。
 

Input
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。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)
 

Output
输出 一行有两个数, 最短距离及其花费。
 

Sample Input
 
 
3 2 1 2 5 6 2 3 4 5 1 3 0 0
 

Sample Output
 
 
9 11
 

Source
                    这道题我竟然在INFINITY这个点wa了许久,例子:#define INFINITY 0x7fffffff    那么INFINITY + any < 0 成为一个极小的数,那么结果自然就是错的
 
/******************************
*
*	acm:   hdu-3790
*
*	title: 最短路径问题
*
*	time : 2014.8.29
*
*******************************/

/*
    思路:            1/2
           例图  ①   ->  ②
                2/2↘    ↙1/2   (距离/花费)
                      ③
         可见,路径相等时,记录花费较少的即可,可直接用Dijkstra算法。

*/

#include <stdio.h>
#include <stdlib.h>

#define MAXVEX 1001

#define MAXEDGE 100000

#define INFINITY 0x3fffffff
//不能0x7fffffff  =>  加上1变成负数 成了一个很小的数

//构造边集
typedef struct Edge
{
    int cost;   //花费
    int lenth;  //长度
} Edge;

Edge edges[MAXVEX][MAXVEX];


void CreateMGraph(int numVertexes, int numEdges)
{
    int i, j;
    int a, b;
    int len, cost;

    for (i = 1; i <= numVertexes; i++)
    {
        for (j = 1; j <= numVertexes; j++)
        {
            edges[i][j].lenth = INFINITY;
            edges[i][j].cost = INFINITY;
        }
    }

    for (i = 0; i < numEdges; i++)
    {
        scanf("%d%d%d%d", &a, &b, &len, &cost);

        if (edges[a][b].lenth > len)
        {
            edges[a][b].lenth = len;
            edges[a][b].cost = cost;

            edges[b][a].lenth = len;
            edges[b][a].cost = cost;
        }
        else if (edges[a][b].lenth == len)
        {
            if (edges[a][b].cost > cost)
            {
                edges[a][b].cost = cost;
                edges[b][a].cost = cost;
            }
        }
    }

  /*        静态数据
            edges[1][2].lenth = 5;
            edges[1][2].cost = 6;

            edges[2][1].lenth = 5;
            edges[2][1].cost = 6;

            edges[2][3].lenth = 4;
            edges[2][3].cost = 5;

            edges[3][2].lenth = 4;
            edges[3][2].cost = 5;
*/
}

typedef int ShortPathTable[MAXVEX];
typedef int Cost[MAXVEX];

//Dijkstra算法
void ShortestPath_Dijkstra(int numVertexes, int numEdges, int v0, int vm, int *len, int *cost)
{
    int v, w, k, min;
    int min_cost;
    ShortPathTable D;
    int final[MAXVEX];  //标记是否求得顶点v0-vm的路径
    Cost C;

    *len = INFINITY;
    *cost = INFINITY;

    for (v = 1; v <= numVertexes; v++)
    {
        final[v] = 0;
        D[v] = edges[v0][v].lenth;
        C[v] = edges[v0][v].cost;
    }

    final[v0] = 1;

    //开始主循环,每次求得v0到某v顶点的最短路径
    for (v = 2; v <= numVertexes; v++)
    {
        min = INFINITY;           //当前所知离v0顶点的最近距离
        min_cost = 0;

        for (w = 1; w <= numVertexes; w++)   //寻找离v0最近的顶点
        {
            if (!final[w] && D[w] < min)
            {
                k = w;
                min = D[w];  //w顶点离v0更近
                min_cost = C[w];
            }
        }

        final[k] = 1;

        if (k == vm)
        {
            *len = D[k];
            *cost = C[k];

            break;
        }

        for (w = 1; w <= numVertexes; w++)
        {
            if (!final[w] && (min+edges[k][w].lenth) < D[w])  //如果经过v顶点的路径比现在这条路径的长度短的话
            {
                //说明找到了更短的路径,修改D[w],C[w]
                D[w] = min + edges[k][w].lenth;
                C[w] = min_cost + edges[k][w].cost;
            }
            else if (!final[w] && (min+edges[k][w].lenth) == D[w])  //如果经过v顶点的路径与现在这条路径相等
            {
                if (min_cost + edges[k][w].cost < C[w])    //取花费较少的值赋值
                {
                    C[w] = min_cost + edges[k][w].cost;
                }
            }
        }
    }

}

int main()
{
    int numVertexes = 3;
    int numEdges = 2;
    int start;
    int end;

    int len;
    int cost;

    while (scanf("%d%d", &numVertexes, &numEdges) && numVertexes != 0 && numVertexes != 0)
    {

        CreateMGraph(numVertexes, numEdges);

        scanf("%d%d", &start, &end);

        ShortestPath_Dijkstra(numVertexes, numEdges, start, end, &len, &cost);

        printf("%d %d\n", len, cost);
    }

    return 0;
}