1030. Travel Plan (30)-PAT甲级真题

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

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分析：这道题是很标准的最短路径加上开销的问题，就是用Dijkstra求出最短的路径，再用一个pre保存每一个节点的前置节点，再dfs遍历运算统计开销的最小值最后得出结果。

#include <cstdio>#include <algorithm>#include <vector>using namespace std;int n , m , s , d;int e[510][510] , dis[510] , cost[510][510];vector<int> pre[510];bool visit[510];const int inf = 99999999;vector<int> path , temppath;int mincost = inf;void dfs (int v){    if (v == s)    {        temppath.push_back (v);        int tempcost = 0;        for (int i = temppath.size () - 1; i > 0; i--)        {            int id = temppath[i] , nextid = temppath[i - 1];            tempcost += cost[id][nextid];        }        if (tempcost < mincost)        {            mincost = tempcost;            path = temppath;        }        temppath.pop_back ();        return;    }    temppath.push_back (v);    for (int i = 0; i < pre[v].size (); i++)        dfs (pre[v][i]);    temppath.pop_back ();}int main (){    fill (e[0] , e[0] + 510 * 510 , inf);    fill (dis , dis + 510 , inf);    scanf_s ("%d%d%d%d" , &n , &m , &s , &d);    for (int i = 0; i < m; i++)    {        int a , b;        scanf_s ("%d%d" , &a , &b);        scanf_s ("%d" , &e[a][b]);        e[b][a] = e[a][b];        scanf_s ("%d" , &cost[a][b]);        cost[b][a] = cost[a][b];    }    pre[s].push_back (s);    dis[s] = 0;    for (int i = 0; i < n; i++)    {        int u = -1 , minn = inf;        for (int j = 0; j < n; j++)        {            if (visit[j] == false && dis[j] < minn)            {                u = j;                minn = j;            }        }        if (u == -1) break;        visit[u] = true;        for (int v = 0; v < n; v++)        {            if (visit[v] == false && e[u][v] != inf)            {                if (dis[v] > dis[u] + e[u][v])                {                    dis[v] = dis[u] + e[u][v];                    pre[v].clear ();                    pre[v].push_back (u);                }                else if (dis[v] == dis[u] + e[u][v])                {                    pre[v].push_back (u);                }            }        }    }    dfs (d);    for (int i = path.size () - 1; i >= 0; i--)        printf ("%d " , path[i]);    printf ("%d %d" , dis[d] , mincost);    return 0;}