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
分析：30分这种题就是考数据结构，题目意思很明确。寻找最短路径，最短路径不唯一，但可通过花费找到唯一的一条。alg:dijkstra(找到所有的最短路径)+dfs(找到其中花费最少的那条)。
code:
#include<iostream> #include<climits>//INT_MAX头文件 #include<vector> using namespace std; const int MAX=INT_MAX; int N,M,S,D; int dist[510][510],cost[510][510]; int visit[510]; vector<int>pre[510]; //dfs选出相同路径中花费最少的 int dfs(int d,int c) { if(d==S) return c; int tmp_a,tmp_b,min; tmp_b=MAX; //选出花费最少 for(int i=0;i<pre[d].size();i++) { tmp_a=dfs(pre[d].at(i),c+cost[d][pre[d].at(i)]); if(tmp_a<tmp_b) { tmp_b=tmp_a; min=pre[d].at(i); } } cost[min][min]=d;//路径链接 return tmp_b; } int main() { //保存记录 fill(cost[0],cost[0]+510*510,MAX); fill(dist[0],dist[0]+510*510,MAX); fill(visit,visit+510,0); cin>>N>>M>>S>>D; int tmp_a,tmp_b; for(int i=0;i<M;i++) { cin>>tmp_a>>tmp_b; cin>>dist[tmp_a][tmp_b]; cin>>cost[tmp_a][tmp_b]; dist[tmp_b][tmp_a]=dist[tmp_a][tmp_b]; cost[tmp_b][tmp_a]=cost[tmp_a][tmp_b]; } //dijkstra找出最短路径 int min=S; dist[min][min]=0; int min_cost=MAX; //dijkstra算法 while(true) { //get min min_cost=MAX; for(int i=0;i<N;i++) { if(visit[i]==0&&dist[i][i]<min_cost) { min_cost=dist[i][i]; min=i; } } if(visit[min]==1) break; dist[min][min]=min_cost; visit[min]=1; //relax 松弛 for(int i=0;i<N;i++) { if(visit[i]==1||dist[min][i]==MAX) continue; if(dist[min][min]+dist[min][i]<dist[i][i]) { dist[i][i]=dist[min][min]+dist[min][i]; pre[i].clear(); pre[i].push_back(min); //最短路径不是一条 }else if(dist[min][min]+dist[min][i]==dist[i][i]) pre[i].push_back(min); } } min_cost=dfs(D,0); int cur=S; while(true) { cout<<cur<<" "; cur=cost[cur][cur]; if(cur==D) break; } cout<<cur<<" "<<dist[D][D]<<" "<<min_cost<<endl; return 0; } 参考 柳婼 の blog 1030. Travel Plan (30)-PAT甲级真题（Dijkstra + DFS，输出路径，边权）