模板2 最短路计数

给出一个 NN 个顶点 MM 条边的无向无权图，顶点编号为 1-N1−N 。问从顶点 11 开始，到其他每个点的最短路有几条。

典型的最短路，加上一个数组代表到达这个点有几个路径

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 2100000;
int head[maxn], dis[maxn], cnt, n, m;
int ans[maxn];
bool vis[maxn];
struct edge
{
    int v, w, next;
}Edge[maxn*3];
struct node
{
    int point, distance;
    node(){}
    node(int _point, int _distance) {point = _point; distance = _distance;}
    bool operator < (const node &other)const
    {
        return distance > other.distance;
    }
};
void add_Edge(int u, int v, int w)
{
    Edge[cnt].v = v;
    Edge[cnt].w = w;
    Edge[cnt].next = head[u];
    head[u] = cnt++;
}
void Dijkstra(int s)
{
    for(int i = 0; i <= n; i++)
        dis[i] = INF;
    memset(ans,0,sizeof(ans));
    ans[s]=1;
    memset(vis, false, sizeof(vis));
    dis[s] = 0;
    priority_queue<node> q;
    q.push(node(s, dis[s]));
    while(!q.empty())
    {
        node now = q.top();
        q.pop();
        if(vis[now.point])
            continue;
        vis[now.point] = true;
        for(int i = head[now.point]; i != -1; i = Edge[i].next)
        {
            int to = Edge[i].v;
            if(dis[to] > dis[now.point] + Edge[i].w)
            {
                dis[to] = dis[now.point] + Edge[i].w;
                ans[to]=ans[now.point];
                q.push(node(to, dis[to]));
            }
            else if(dis[to]==dis[now.point]+Edge[i].w)
            {
                ans[to]=(ans[to]+ans[now.point])%100003;
            }
        }
    }
}
int main()
{
    int u, v, w;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        if(n == 0 && m == 0)
            break;
        cnt = 0;
        memset(head, -1, sizeof(head));
        while(m--)
        {
            scanf("%d%d", &u, &v);
            w=1;
            add_Edge(u, v, w);
            add_Edge(v, u, w);
        }
        Dijkstra(1);
        for(int i=1;i<=n;i++)
            cout<<ans[i]<<endl;
    }
    return 0;
}

若是无向图，则一样操作