07-图6 旅游规划 (25分)

题目描述

有了一张自驾旅游路线图，你会知道城市间的高速公路长度、以及该公路要收取的过路费。现在需要你写一个程序，帮助前来咨询的游客找一条出发地和目的地之间的最短路径。如果有若干条路径都是最短的，那么需要输出最便宜的一条路径。

输入格式:

输入说明：输入数据的第1行给出4个正整数N、M、S、D，其中N（2≤N≤500）是城市的个数，顺便假设城市的编号为0~(N−1)；M是高速公路的条数；S是出发地的城市编号；D是目的地的城市编号。随后的M行中，每行给出一条高速公路的信息，分别是：城市1、城市2、高速公路长度、收费额，中间用空格分开，数字均为整数且不超过500。输入保证解的存在。

输出格式:

在一行里输出路径的长度和收费总额，数字间以空格分隔，输出结尾不能有多余空格。

输入样例:

输出样例:

3 40

解题思路

根据题意，可以看出这是一个单源有权图最短路径问题，使用Dijkstra算法解决，需要注意的是本题最短路径的含义是先比较长度，再比较花费。

代码

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

#define MAXSIZE 500
#define INFINITY 600

typedef struct {
    int length;
    int pay;
} Edge;

typedef struct {
    int vertexCount;
    int edgeCount;
    Edge matrix[MAXSIZE][MAXSIZE];
} Graph, *PGraph;

PGraph createGraph(int N, int M);
void initDistance(Edge dis[], PGraph graph, int S);
void dijkstra(Edge dis[], PGraph graph, int S);
int getNext(Edge dis[], int N);

visited[MAXSIZE] = {0};

int main() {
    int N, M, S, D;
    scanf("%d %d %d %d", &N, &M, &S, &D);
    PGraph graph = createGraph(N, M);
    Edge dis[MAXSIZE];              //记录路径长度与花费
    initDistance(dis, graph, S);
    dijkstra(dis, graph, S);
    printf("%d %d\n", dis[D].length, dis[D].pay);
    return 0;
}

PGraph createGraph(int N, int M) {
    PGraph graph = (PGraph) malloc(sizeof(Graph));
    graph->vertexCount = N;
    graph->edgeCount = M;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            graph->matrix[i][j].length = INFINITY;
            graph->matrix[i][j].pay = INFINITY;
        }
    }
    for (int i = 0; i < M; i++) {
        int x, y, len, pay;
        scanf("%d %d %d %d", &x, &y, &len, &pay);
        graph->matrix[x][y].length = len;
        graph->matrix[x][y].pay = pay;
        graph->matrix[y][x].length = len;
        graph->matrix[y][x].pay = pay;
    }
    return graph;
}

void initDistance(Edge dis[], PGraph graph, int S) {
    for (int i = 0; i < graph->vertexCount; i++) {
        if (i == S) {
            dis[i].length = 0;
            dis[i].pay = 0;
        } else {
            dis[i].length = graph->matrix[S][i].length;
            dis[i].pay = graph->matrix[S][i].pay;
        }
    }
}

void dijkstra(Edge dis[], PGraph graph, int S) {
    visited[S] = 1;         //将起点标记为已访问
    int N = graph->vertexCount;
    while (1) {
        int next = getNext(dis, N);
        if (next == -1) break;
        visited[next] = 1;
        for (int i = 0; i < N; i++) {
            //对next每个未访问的邻接点
            if (graph->matrix[next][i].length != INFINITY && !visited[i]) {
                if (dis[i].length > dis[next].length + graph->matrix[next][i].length) {
                    dis[i].length = dis[next].length + graph->matrix[next][i].length;
                    dis[i].pay = dis[next].pay + graph->matrix[next][i].pay;
                } else if (dis[i].length == dis[next].length + graph->matrix[next][i].length) {
                    if (dis[i].pay > dis[next].pay + graph->matrix[next][i].pay) {
                        dis[i].length = dis[next].length + graph->matrix[next][i].length;
                        dis[i].pay = dis[next].pay + graph->matrix[next][i].pay;
                    }
                }
            }
        }
    }   
}

int getNext(Edge dis[], int N) {
    int next = -1;
    int minLength = INFINITY, minPay = INFINITY;
    for (int i = 0; i < N; i++) {
        if (!visited[i] && minLength >= dis[i].length) {
            if (minLength > dis[i].length) {
                minLength = dis[i].length;
                minPay = dis[i].pay;
                next = i;
            } else if (minLength == dis[i].length) {
                if (minPay > dis[i].pay) {
                    minLength = dis[i].length;
                    minPay = dis[i].pay;
                    next = i;
                }
            }
        }
    }
    return next;
}

来源：https://www.cnblogs.com/AndyHY-Notes/p/12631241.html

标签

matrix

dis