1030. Travel Plan (30)

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本文介绍了一个旅行者如何通过编程寻找从起点到终点的最短路径，并在有多条最短路径时选择成本最低的一条。使用深度优先搜索算法来实现这一目标。

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

Sample Output

0 2 3 3 40

思路：用深度优先遍历来记录最短路径上的各个节点；因为可能存在相同的最短路径，增加额外的一个节点记录当前最短路径的信息，于找到的相比较就OK！

参考代码：

#include <iostream>
#include <fstream>
using namespace std;
struct Node{
    int length;
    int cost;
};
struct checkI{
    int totalLength;
    int totalCost;
    int road[501];
    int totalCount;
};
const int INF = 99999999;
int N,M,start,dest;
int visited[501];
int curCost = 0;
int curDis = 0;
int flag = 0;
int path[500];
Node edge[501][501];
checkI ch;

int DFS(int start){
    path[flag++] = start;
    if(start==dest){
        if(curDis<ch.totalLength){                  //此次遍历的是最短路径，记录在ch变量中
            ch.totalLength = curDis;
            ch.totalCost = curCost;
            ch.totalCount = flag;
            for(int i=0;i<flag;i++){
                ch.road[i] = path[i];                    //记录最短路径上的各个节点
            }
        }else{
            if(curDis == ch.totalLength){
                if(curCost<ch.totalCost){
                    ch.totalLength = curDis;
                    ch.totalCost = curCost;
                    ch.totalCount = flag;
                    for(int i=0;i<flag;i++){
                        ch.road[i] = path[i];   //记录最短路径上的各个节点
                    }
                }
            }
        }
    }else{
        visited[start] = 1;
        int i;
        for(i=0;i<N;i++){
            if(!visited[i] && edge[start][i].length<INF){
                curDis += edge[start][i].length;
                curCost += edge[start][i].cost;
                DFS(i);
                flag--;
                curDis -= edge[start][i].length;
                curCost -= edge[start][i].cost;
            }
        }
    }
}

int main()
{
    cin>>N>>M>>start>>dest;
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            if(i==j){
                edge[i][j].length = 0;
            }else{
                edge[i][j].length = INF;
                edge[j][i].length = INF;
            }
        }
    }
    for(int i=0;i<M;i++){
        int a,b,c,d;
        cin>>a>>b>>c>>d;
        edge[a][b].length = c;
        edge[a][b].cost = d;
        edge[b][a].length = c;
        edge[b][a].cost = d;
    }
    ch.totalLength = INF;
    DFS(start);
    for(int i=0;i<ch.totalCount;i++)
        cout<<ch.road[i]<<" ";
    cout<<ch.totalLength<<" "<<ch.totalCost;
    return 0;
}