问题描述
我正在尝试在C中使用BackTracking解决以下问题，但我不知道如何从这里继续...
问题是：从城市A到城市B的道路长度相比而言并不相同到从城市B到城市A的道路。
从同一来源城市返回到它的路（直接）为0。
克里斯注意到从A到B的最短道路并不总是直接两个城市之间的一个。
您将需要帮助Chris查找最短路径。
编写一个函数，该函数在给定矩阵NxN的情况下检查最短地图，该矩阵存储道路长度的值。
注意：N定义为4。
示例：
从0到3的最短路径1给城市0，然后3，然后给1，如果给定以下矩阵：
0 5 2 2
1 0 1 1
1 2 0 1
1 1 2 0
她的代码：
int ShortestPath（int SourceCity，int DestinationCity，int Distance [] [N ]，bool Chosen [] [N]）
{
int Path = 0;
if（SourceCity == DestinationCity）
{
Distance [SourceCity] [DestinationCity] = true;
返回0；
}
for（int i = 0; i {
for（int j = 0; j {
路径+ =距离[i] [j];
if（！Chosen [i] [j]）
{
Chosen [i] [j] = true;
ShortestPath（i，DestinationCity，Distance，Chosen）;
}
}
}
if（Path> = Distance [SourceCity] [DestinationCity]）
{
Chosen [SourceCity，DestinationCity] = false;
return Distance [SourceCity] [DestinationCity];
}
}
注意：选择的矩阵表示我是否选择了特定的道路与否（其初始值全为假）
以下是建议的解决方案。这是Dijkstra算法的实现，它找到了最短的路径。
#include< stdio.h>
#include< stdlib.h>
#include< limits.h>
#include< stdbool.h>
//节点数
#define N 4
// d [i] [j]从i到j的距离矩阵
int d [N] [N] = {
{0，5，2，2}，
{1，0，1，1}，
{1,2,0,1}，
{1，1，2，0}，
};
// shortestPath使用距离矩阵d在路径中存储从起始节点到
//最终节点的最短路径。它返回路径中
//节点的数量。
int shortestPath（int d [N] [N]，int start，int final，int path [N]）{
//节点i之前的节点
int prev [N];
//初始化到节点i的距离，开始为无限的
int dist [N];
for（int i = 0; i< N; i ++）
dist [i] = INT_MAX;
dist [start] = 0;
//初始化已完成节点的列表
bool done [N];
for（int i = 0; i< N; i ++）
done [i] = false;
int nDone = 0;
//虽然我们还没有完成所有节点
，但是（nDone< N）{
//找到尚未完成的节点，以最小的距离开始节点
int minDist = INT_MAX;
int n; //对于（int i = 0; i< N; i ++）具有最小距离
的节点if（！done [i]&& dist [i]< minDist）
minDist = dist [n = i];
done [n] = true;
nDone ++;
//如果（n == final）
中断，我们可以在完成最后一个节点
后停止。
//对于每个节点j ...
for（int j = 0; j< N; j ++）{
//如果节点j尚未完成，
//，如果（（done [j]&& dist [j]> dist [n] + d [n] [j ]）{
//设置新的最短距离
dist [j] = dist [n] + d [n] [j];
//将节点n设置为节点j之前的节点
prev [j] = n;
}
}
}
//获取路径[start，...，final]
int j = N;
for（int i = final; i！= start; i = prev [i]）
path [-j] = i;
path [-j] =开始；
如果（j == 0）
返回N;
int n = N-j;
for（int i = 0; i< n; i ++，j ++）
path [i] = path [j];
返回n；
}
int main（）{
int path [N];
int n = shortestPath（d，0，1，path）;
printf（ path：％d，path [0]）;
for（int i = 1; i< n; i ++）
printf（->％d，path [i]）;
printf（ \n）;
返回0；
}
输出
路径：0-> 3-> 1
I am trying to solve the following question using BackTracking in C but I don't know how to continue from here...
The question is:
Chris is planning to travel in a country with N cities. He will get help from a matrix NxN that the cell (I,J) represents the length of the road from city I to City J.The length of the road from City A to city B is not the same when compared to the road from City B to City A.The Road From the same source city back to it (directly) is 0.Chris noticed that the shortest road From A to B isn't alway the direct one between both Cities.You will need to help Chris Finding the shortest path.Write a function that checks for the shortest map given a matrix NxN which stores the values of the roads lengths.Note: N is defined as 4.
Example:
The Shortest path from 0 to 1 is going to City 0 then 3 then 1 if given the following matrix:
0 5 2 2
1 0 1 1
1 2 0 1
1 1 2 0
Her's my code:
int ShortestPath (int SourceCity, int DestinationCity, int Distance [][N], bool Chosen[][N])
{
int Path=0;
if (SourceCity==DestinationCity)
{
Distance[SourceCity][DestinationCity]=true;
return 0;
}
for (int i=0;i<N;i++)
{
for (int j=0;j<N;j++)
{
Path += Distance[i][j];
if (!Chosen[i][j])
{
Chosen[i][j] = true;
ShortestPath(i, DestinationCity, Distance, Chosen);
}
}
}
if (Path>=Distance[SourceCity][DestinationCity])
{
Chosen[SourceCity,DestinationCity]=false;
return Distance[SourceCity][DestinationCity];
}
}
Note: the matrix chosen indicated wether I chose a specific road or not (Its initial values are all false)
Here is a suggested solution. It is an implementation of the Dijkstra algorithm finding the shortest path.
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>
// number of nodes
#define N 4
// distance matrix from i to j for d[i][j]
int d[N][N] = {
{0, 5, 2, 2},
{1, 0, 1, 1},
{1, 2, 0, 1},
{1, 1, 2, 0},
};
// shortestPath stores in path the shortest path from start node to
// final node using the distance matrix d. It returns the number of
// nodes in the path.
int shortestPath(int d[N][N], int start, int final, int path[N]){
// node previous to node i
int prev[N];
// initialize distance from node i to start as infinite
int dist[N];
for (int i = 0; i < N; i++)
dist[i] = INT_MAX;
dist[start] = 0;
// initialize list of nodes done
bool done[N];
for (int i = 0; i < N; i++)
done[i] = false;
int nDone = 0;
// while we haven’t done all nodes
while (nDone < N) {
// find not yet done node with minimal distance to start node
int minDist = INT_MAX;
int n; // node with minimum distance
for (int i = 0; i < N; i++)
if (!done[i] && dist[i] < minDist)
minDist = dist[n = i];
done[n] = true;
nDone++;
// we can stop when final node is done
if (n == final)
break;
// for every node j...
for (int j = 0; j < N; j++) {
// if node j is not yet done,
// and distance from start to j through n is smaller to known
if (!done[j] && dist[j] > dist[n] + d[n][j]) {
// set new shortest distance
dist[j] = dist[n] + d[n][j];
// set node n as previous to node j
prev[j] = n;
}
}
}
// get path [start, ..., final]
int j = N;
for (int i = final; i != start; i = prev[i])
path[--j] = i;
path[--j] = start;
if (j == 0)
return N;
int n = N-j;
for (int i = 0; i < n; i++, j++)
path[i] = path[j];
return n;
}
int main() {
int path[N];
int n = shortestPath(d, 0, 1, path);
printf("path: %d", path[0]);
for (int i = 1; i < n; i++)
printf("->%d", path[i]);
printf("\n");
return 0;
}
It outputs
path: 0->3->1
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