（算法）無向圖最短路徑的數目

題目：

給定如下圖所示的無向連通圖，假定圖中所有邊的權值都為1；
顯然，從源點A到終點T的最短路徑有多條，求不同的最短路徑的數目。
注：兩條路徑中有任意結點不同或者結點順序不同，都稱為不同的路徑。

 

 

思路：

給定的圖中，邊權相等且非負，Dijkstra最短路徑算法退化為BFS廣度優先搜索。實現過程中可以使用隊列。
計算到某結點最短路徑條數，只需計算與該結點相鄰的結點的最短路徑值和最短路徑條數，把最短路徑值最小且相等的最短路徑條數加起來即可。

答案：12

代碼：

#include <iostream>
#include <queue>
#include <string.h>

using namespace std;

const int N=16;

int calNumOfPath(int G[N][N]){
    int stepNum[N]; // how many steps to reach i
    int pathNum[N]; // how many paths can reach i
    bool visited[N];
    memset(stepNum,0,N*sizeof(int));
    memset(pathNum,0,N*sizeof(int));
    memset(visited,false,N*sizeof(bool));
    stepNum[0]=0;
    pathNum[0]=1;

    queue<int> q;
    q.push(0);

    while(!q.empty()){
        int node=q.front();
        q.pop();
        visited[node]=true;
        int s=stepNum[node]+1;
        for(int i=0;i<N;i++){
            if(i!=node && !visited[i] && G[node][i]==1){
                if(stepNum[i]==0 || pathNum[i]>s){
                    stepNum[i]=s;
                    pathNum[i]=pathNum[node];
                    q.push(i);
                }
                else if(stepNum[i]==s){
                    pathNum[i]=pathNum[i]+pathNum[node];
                }
            }
        }
    }
    return pathNum[N-1];
}

int main()
{
    int G[16][16]={
    {0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0},
    {1,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0},
    {0,1,0,1,0,0,1,0,0,0,0,0,0,0,0,0},
    {0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0},
    {1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0},
    {0,1,0,0,1,0,1,0,0,1,0,0,0,0,0,0},
    {0,0,1,0,0,1,0,1,0,0,1,0,0,0,0,0},
    {0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0},
    {0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0},
    {0,0,0,0,0,1,0,0,1,0,1,0,0,1,0,0},
    {0,0,0,0,0,0,1,0,0,1,0,1,0,0,1,0},
    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1},
    {0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0},
    {0,0,0,0,0,0,0,0,0,1,0,0,1,0,1,0},
    {0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,1},
    {0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0}};
    cout << calNumOfPath(G) << endl;
    return 0;
}