pat 1030

1030. Travel Plan (30)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 30 1 1 201 3 2 300 3 4 100 2 2 202 3 1 20

Sample Output

0 2 3 3 40

//该题重点是存储最短路径的方法。path[j] = p;表示到达城市j的最短路径中前一个城市是p#include <stdio.h>#include <string.h>#include <vector>using namespace std;#define INF 1 << 20int vis[510], path[510], dist[510], cost[510];int map[510][510], mapCost[510][510];int n, m, s, d;vector<int> Q;void dij(){int i, j;for(i = 0; i < n; i++)dist[i] = INF;dist[s] = 0;cost[s] = 0;for(i = 0; i < n; i++){int min = INF, index = -1;for(j = 0; j < n; j++){if(vis[j] == 0 && dist[j] < min){min = dist[j];index = j;}}if(min == INF || index == d) return;vis[index] = 1;for(j = 0; j < n; j++){if(vis[j] == 0){int tmp = dist[index] + map[index][j];if(tmp < dist[j]){dist[j] = tmp;path[j] = index; //到达城市j的最短路径上的前一个城市是城市indexcost[j] = cost[index] + mapCost[index][j]; //cost[j]表示从起始的城市s到城市j的最小花费}else if(tmp == dist[j]){if(cost[j] > cost[index] + mapCost[index][j]){//如果这条路的花费高path[j] = index; //修改原先到达城市j的前一个城市是path[j]的路径，换成到达城市j的前一个城市是index的路径cost[j] = cost[index] + mapCost[index][j];}}}}}}int main(){int i, j, c1, c2, di, co;while(scanf("%d%d%d%d", &n, &m, &s, &d) != EOF){for(i = 0; i <= n; i++)for(j = 0; j <= n; j++)map[i][j] = map[j][i] = INF;for(i = 1; i <= m; i++){scanf("%d%d%d%d", &c1, &c2, &di, &co);map[c1][c2] = map[c2][c1] = di;mapCost[c1][c2] = mapCost[c2][c1] = co;}memset(vis, 0, sizeof(vis));dij();Q.clear();int tmp = d;while(d != s){Q.push_back(d);d = path[d];}Q.push_back(s);for(i = Q.size() - 1; i >= 0; i--)printf("%d ", Q[i]);printf("%d %d\n", dist[tmp], cost[tmp]);}return 0;}