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.

const int MAXV = 510;//最大顶点数 const int INF = 1000000000;//无穷大

//n为顶点数，m为边数，st和ed为起点和终点 //G为距离矩阵，cost为花费矩阵 //d[]记录最短距离，c[]记录最小花费 int n, m, st, ed, G[MAXV][MAXV], cost[MAXV][MAXV]; int d[MAXV], c[MAXV], pre[MAXV]; bool vis[MAXV] = {false};//vis[i]表示顶点i已访问，初值均为false

void Dijkstra(int s){//s为起点 fill(d, d + MAXV, INF);//fill函数将整个d数组赋为INF fill(c, c + MAXV, INF);//fill函数将整个c数组赋为INF d[s] = 0; c[s] = 0; for (int i = 0; i < n; i++) { int u = -1, MIN = INF; for (int j = 0; j < n; j++) { if (vis[j] == false && d[j] < MIN) { MIN = d[j]; u = j; } } if (u == -1) return; vis[u] = true; for (int v = 0; v < n; v++) { if (vis[v] == false && G[u][v] != INF) { if (d[u] + G[u][v] < d[v]) { d[v] = d[u] + G[u][v]; c[v] = c[u] + cost[u][v]; pre[v] = u; }else if(d[u] + G[u][v] == d[v] && c[u] + cost[u][v] < c[v]){ c[v] = c[u] + cost[u][v]; pre[v] = u; } } } } }

void DFS(int v){//打印路径 if (v == st) {//递归终点 printf("%d ", v); return; } DFS(pre[v]); printf("%d ", v); }

int main(){ scanf("%d%d%d%d", &n, &m, &st, &ed); int u, v; fill(G[0], G[0] + MAXV * MAXV, INF);//初始化图 for (int i = 0; i < m; i++) { scanf("%d%d", &u, &v); scanf("%d%d", &G[u][v], &cost[u][v]); G[v][u] = G[u][v]; cost[v][u] = cost[u][v]; } Dijkstra(st); DFS(ed); printf("%d %d\n", d[ed], c[ed]); return 0; }