查找乘坐火车到达目的地的最低费用
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。
输入输出
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.
算法
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示例
#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