（算法）无向图最短路径的数目

题目：

给定如下图所示的无向连通图，假定图中所有边的权值都为1；
显然，从源点A到终点T的最短路径有多条，求不同的最短路径的数目。
注：两条路径中有任意结点不同或者结点顺序不同，都称为不同的路径。

思路：

给定的图中，边权相等且非负，Dijkstra最短路径算法退化为BFS广度优先搜索。实现过程中可以使用队列。
计算到某结点最短路径条数，只需计算与该结点相邻的结点的最短路径值和最短路径条数，把最短路径值最小且相等的最短路径条数加起来即可。

答案：12

代码：

#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;
}