对于此问题,旅途中有N个站点。车辆从0号站开始行驶至N-1。在表格中,列出了所有两个车站的票价。我们必须找到使用这些给定成本到达目的地的最低成本。
Input: The cost matrix of the journey. 0 15 80 90 ∞ 0 40 50 ∞ ∞ 0 70 ∞ ∞ ∞ 0 Output: The minimum cost is 65. At first go to the destination 1 from 0. (cost 15), then 1 to 4 (cost 50). So total cost 65.
findMinCost(cost)
输入-从每个来源到每个目的地的成本矩阵。
输出-找到到达目的地的最低费用。
Begin define array costLoc, whose size is same as sumber of locations, fill costLoc with ∞. n := number of locations costLoc[0] := 0 for each source i to each destination j, do if costLoc[j] > costLoc[i] + cost[i, j], then costLoc[j] := costLoc[i] + cost[i, j] done return costLoc[n-1] End
#include<iostream> #define INF INT_MAX #define NODE 4 using namespace std; int cost[NODE][NODE] = { {0, 15, 80, 90}, {INF, 0, 40, 50}, {INF, INF, 0, 70}, {INF, INF, INF, 0} }; int findMinCost() { //find smallest possible cost to reach destination int costStation[NODE]; //store cost to reach any station from 0 for (int i=0; i<NODE; i++) costStation[i] = INF; //initially all are infinity costStation[0] = 0; //cost for station 0 is 0 as it is starting point for (int i=0; i<NODE; i++) for (int j=i+1; j<NODE; j++) if (costStation[j] > costStation[i] + cost[i][j]) //find intermediate station for min cost costStation[j] = costStation[i] + cost[i][j]; return costStation[NODE-1]; } int main() { cout << "The Minimum cost to reach station " << NODE << " is " << findMinCost() << endl; return 0; }
输出结果
The Minimum cost to reach station 4 is 65