FloydWarshall多(所有)源点的最短路径实现(带路径打印)

一、介绍和算法原理:

https://www.cs.usfca.edu/~galles/visualization/Floyd.html

二、核心代码:

dist[i][j]表示从i到j的最短距离，parent[i][j]表示源点为i时，j的双亲是谁，一开始我都初始化parent[i][j] = i
下面的是这个算法的核心代码，书上讲的挺好的，dist^(k)[i][j]表示从i到j，中间由序号不大于k的顶点序号(可能是多个)作为中转点时,i与j的最短路径。

 for(int k = 0; k < V; k++){for(int i = 0; i < V; i++){for(int j = 0; j < V; j++){if(dist[i][k] + dist[k][j] < dist[i][j]){dist[i][j] = dist[i][k] + dist[k][j]; parents[i][j] = parents[k][j];  }}}}

1. 这里外循环k=0,里面两层循环尝试更新所有i-->j以k=0为中间节点时的最短距离。可能会出现这些情况，1-->0-->5,2-->0-->3,3-->0-->1,（0-->0-->5这种就没意义)
2. 当k=1时，尝试所有i->1->j路径，因为上一步k=0的时候检查过1->0->(0,1,2...n-1)会不会让1到其他各点的距离会不会变短，所以这个i->1->j是在上一步的基础上的，即2-->1--->5可能代表的这条路径2-->1-->0-->5，3-->1--->4可能代表的这条路径3-->1-->0-->4,当然0-->1--->4也是可能的。
3. k=2时，尝试以2为中间点(事实上是尝试把2加到中间点序列),即0-->2-->5可能代表着0-->1-->2-->5,3-->2-->6可能代表着3-->1-->2-->0---->6,但是不会出现3-->5--->1--->6这种情况，因为5还没做为中间结点去检查过。
4…
5…
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在内外循环结束之后，所有结点都尝试被当做中间结点加入i与j的路径中去了，所以得到dist[i][j]就i到j的最短路径了。

二、全部代码:

#include<iostream>
#include<vector>using namespace std;#define INF 999999    //不可能的权值
#define V 6         //顶点数

void printSolution(int dist[][V],int parents[][V],int graph[][V])
{cout<<"打印任一对顶点间的最短路径,"<<"若weight为INF和dist[i][j]为INF则表示i->j无路径"<<endl;  for(int i = 0; i < V; i++){cout<<"------从"<<i<<"出发到各顶点的最短路径:-------"<<endl; for(int j = 0; j < V; j++){cout<<parents[i][j]<<" --> "<<j<<" \t weight is :";int parent = parents[i][j];if (graph[parent][j] == INF)  cout<<"INF"<<"     ";  elsecout<<graph[parent][j]<<"     ";cout<<" dist["<<i<<"]["<<j<<"] is ";if (dist[i][j] == INF)  cout<<"INF"<<endl;  elsecout<<dist[i][j]<<endl;   }cout<<endl;}
}

void floydWarshall(int graph[][V])
{int dist[V][V]; int parents[V][V]; //parent[i][0...V-1]是从第i个元素//出发的双亲数组，用来记录路径 for(int i = 0; i < V; i++){for(int j = 0; j < V; j++){dist[i][j] = graph[i][j];//从i出发时将i视为其他顶点的双亲,若后面变了,说明加了中转顶点 parents[i][j] = i;    } }for(int k = 0; k < V; k++){for(int i = 0; i < V; i++){for(int j = 0; j < V; j++){if(dist[i][k] + dist[k][j] < dist[i][j]){dist[i][j] = dist[i][k] + dist[k][j]; parents[i][j] = parents[k][j];   }}}}printSolution(dist,parents,graph);
}

int main()
{int graph[V][V] = { {0,   INF,  10, INF, 30,  100},  {INF,   0,   5, INF, INF, INF},  {INF, INF,   0,  50, INF,   1},  {INF, INF, INF,   0, INF,  10}, {INF, INF, INF,  20,   0,  60},{INF, INF, INF,   0, INF,   0}, }; floydWarshall(graph);return 0;
}

输出：

打印任一对顶点间的最短路径,若weight为INF和dist[i][j]为INF则表示i->j无路径
------从0出发到各顶点的最短路径:-------
0 --> 0          weight is :0      dist[0][0] is 0
0 --> 1          weight is :INF      dist[0][1] is INF
0 --> 2          weight is :10      dist[0][2] is 10
5 --> 3          weight is :0      dist[0][3] is 11
0 --> 4          weight is :30      dist[0][4] is 30
2 --> 5          weight is :1      dist[0][5] is 11------从1出发到各顶点的最短路径:-------
1 --> 0          weight is :INF      dist[1][0] is INF
1 --> 1          weight is :0      dist[1][1] is 0
1 --> 2          weight is :5      dist[1][2] is 5
5 --> 3          weight is :0      dist[1][3] is 6
1 --> 4          weight is :INF      dist[1][4] is INF
2 --> 5          weight is :1      dist[1][5] is 6------从2出发到各顶点的最短路径:-------
2 --> 0          weight is :INF      dist[2][0] is INF
2 --> 1          weight is :INF      dist[2][1] is INF
2 --> 2          weight is :0      dist[2][2] is 0
5 --> 3          weight is :0      dist[2][3] is 1
2 --> 4          weight is :INF      dist[2][4] is INF
2 --> 5          weight is :1      dist[2][5] is 1------从3出发到各顶点的最短路径:-------
3 --> 0          weight is :INF      dist[3][0] is INF
3 --> 1          weight is :INF      dist[3][1] is INF
3 --> 2          weight is :INF      dist[3][2] is INF
3 --> 3          weight is :0      dist[3][3] is 0
3 --> 4          weight is :INF      dist[3][4] is INF
3 --> 5          weight is :10      dist[3][5] is 10------从4出发到各顶点的最短路径:-------
4 --> 0          weight is :INF      dist[4][0] is INF
4 --> 1          weight is :INF      dist[4][1] is INF
4 --> 2          weight is :INF      dist[4][2] is INF
4 --> 3          weight is :20      dist[4][3] is 20
4 --> 4          weight is :0      dist[4][4] is 0
3 --> 5          weight is :10      dist[4][5] is 30------从5出发到各顶点的最短路径:-------
5 --> 0          weight is :INF      dist[5][0] is INF
5 --> 1          weight is :INF      dist[5][1] is INF
5 --> 2          weight is :INF      dist[5][2] is INF
5 --> 3          weight is :0      dist[5][3] is 0
5 --> 4          weight is :INF      dist[5][4] is INF
5 --> 5          weight is :0      dist[5][5] is 0--------------------------------
Process exited with return value 0
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