1030. Travel Plan (30)

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本文介绍了一种使用深度优先搜索算法解决旅行者地图中的最短路径问题的方法，包括输入输出规范和代码实现。算法适用于有限城市数量的情况，并确保在存在多种最短路径时选择成本最低的路径。

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原题地址：http://www.patest.cn/contests/pat-a-practise/1030

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

解题思路：dfs回溯即可。

代码如下：

#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <vector>
#define INF 10000000;
using namespace std;

int G[505][505];
int C[505][505];
int vis[505];
int N,M,S,D;

int minDist;
int minCost;
int dist;
int cost;
vector<int> route;
vector<int> tmp;

void dfs_m(int s){
    vis[s] = 1;
    tmp.push_back(s);
    for(int i = 0; i < N; i++){
        if(!vis[i] && G[s][i]){
            dist += G[s][i];
            cost += C[s][i];
            if(dist > minDist){
                dist -= G[s][i];
                cost -= C[s][i];
                continue;
            }
            if(i == D){
                tmp.push_back(i);
                if(dist < minDist){
                    minDist = dist;
                    minCost = cost;
                    route.clear();
                    route.assign(tmp.begin(),tmp.end());
                }else if(dist == minDist){
                    if(cost < minCost){
                        minCost = cost;
                        route.clear();
                        route.assign(tmp.begin(),tmp.end());
                    }
                }

                tmp.pop_back();
                dist -= G[s][i];
                cost -= C[s][i];
                continue;
            }
            dfs_m(i);
            dist -= G[s][i];
            cost -= C[s][i];
        }
    }
    tmp.pop_back();
    vis[s] = 0;
}

int main(){
    freopen("1030.txt","r",stdin);
    scanf("%d%d%d%d",&N,&M,&S,&D);
    memset(G,0,sizeof(G));
    memset(C,0,sizeof(C));
    for(int i = 0;i < M; i++){
        int f,t,d,c;
        scanf("%d%d%d%d",&f,&t,&d,&c);
        G[f][t] = G[t][f] = d;
        C[f][t] = C[t][f] = c;
//      printf("%d-%d: %d : %d\n", f,t,G[f][t],C[f][t]);
    }
    memset(vis,0,sizeof(vis));
    dist = 0;
    cost = 0;
    minDist = INF;
    minCost = INF;
    dfs_m(S);
    for(int i = 0; i < route.size();i++)
        printf("%d ",route[i]);
    printf("%d %d",minDist,minCost);
    return 0;
}