1030. Travel Plan (30)

A traveler’s map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (<=500) is the number of cities (and hence the cities are numbered from 0 to N-1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

City1 City2 Distance Cost

where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:

For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input
4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20

Sample Output
0 2 3 3 40

算法解析：考察点明确，就是单源最短路问题，用dijkstra算法就能解决，只需要做一点改动。

#include <stdio.h>#include <stdlib.h>#define INF 9999999//gdist存储每个城市之间的距离，gcost存储每个城市之间的路费，dist存储从起点到各点的最短距离，//cost类似，path存储路劲，visited表示该城市是否被使用。int gdist[500][500], gcost[500][500], dist[500], cost[500], path[500], visited[500];//初始化void init(int n, int v){    int i;    for(i = 0; i < n; i++)    {        visited[i] = 0;        dist[i] = gdist[v][i];        cost[i] = gcost[v][i];        if(dist[i] < INF) //path[i]=v，表示i的上一个顶点是v。            path[i] = v;        else            path[i] = -1; //-1表示i是根节点。    }}void solve(int n, int v, int w){    int i, j;    visited[v] = 1;    dist[v] = cost[v] = 0;    for(j = 0; j < n; j++)    {        int maxDist = INF, u = v;        for(i = 0; i < n; i++) //选出离起点最近的点。        {            if(!visited[i] && dist[i] < maxDist)            {                u = i;                maxDist = dist[i];            }        }        visited[u] = 1;        for(i = 0; i < n; i++)        {            if(!visited[i] && gdist[u][i] < INF)            {                int newDist = dist[u] + gdist[u][i];                int newCost = cost[u] + gcost[u][i];                if(newDist < dist[i])                {                    dist[i] = newDist;                    cost[i] = newCost;                    path[i] = u;                }                else if(newDist == dist[i] && newCost < cost[i]) //距离相等就比较路费                {                    cost[i] = newCost;                    path[i] = u;                }            }        }    }    int stack[500], top = -1, tmp = w; //创建堆栈，反向输出。    while(tmp > -1)    {        stack[++top] = tmp;        tmp = path[tmp];    }    for(i = top; i >= 0; i--)        printf("%d ", stack[i]);    printf("%d %d", dist[w], cost[w]);}int main(){    int n, m, v, w, i, j, v1, v2, t1, t2;    scanf("%d %d %d %d", &n, &m, &v, &w);    for(i = 0; i < n; i++)        for(j = 0; j < n; j++)            gdist[i][j] = gcost[i][j] = INF;    for(i = 0; i < m; i++)    {        scanf("%d %d %d %d", &v1, &v2, &t1, &t2);        gdist[v1][v2] = gdist[v2][v1] = t1;        gcost[v1][v2] = gcost[v2][v1] = t2;    }    init(n, v);    solve(n, v, w);    return 0;}