最短路计数

基本思路：正常dijkstra+记录路径

注意：路径求模

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int Max_N=1000010,Max_M=2000010,inf=2147432637;
typedef pair<int,int>node;
struct Edge
{
    int next,dis,to;
} E[Max_M*2];//注意是无向图
int Head[Max_N],num_Edge,Dis[Max_N],vis[Max_N],Ans[Max_N];
int N,M;
inline int Read(void)//快读
{
    int w=0,x=0;
    char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=x*10+(ch^48),ch=getchar();
    return w?-x:x;

}
inline void Add_Edge(int from,int to,int dis)//建边
{
    E[++num_Edge].dis=dis;
    E[num_Edge].next =Head[from];
    E[num_Edge].to =to;
    Head[from]=num_Edge;
}
inline void Dijkstra()
{
    priority_queue< node,vector<node>,greater<node> >Q;
    for(int i=2; i<=N; ++i) Dis[i]=inf;
    Ans[1]=1;
    Q.push(make_pair(0,1)) ;
    while(!Q.empty())
    {
        int u=Q.top() .second;
        Q.pop() ;
        if(vis[u])continue;
        else vis[u]=1;
        for(int i=Head[u]; i; i=E[i].next )
        {
            int v=E[i].to;
            if(Dis[v]==Dis[u]+E[i].dis )//累计最短路条数 
                Ans[v]=(Ans[v]+Ans[u])%100003;
            else if(Dis[v]>Dis[u]+E[i].dis )//如果找到更短路，重新统计最短路条数 
            {
                Ans[v]=Ans[u]%100003;
                Dis[v]=Dis[u]+E[i].dis ;
                if(!vis[v]) Q.push(make_pair(Dis[v],v)) ;
            }
        }


    }
}
int main(void)
{
    N=Read(),M=Read();
    for(int i=1; i<=M; ++i)
    {
        int u=Read(),v=Read();
        Add_Edge(u,v,1);
        Add_Edge(v,u,1);
    }
    Dijkstra();

    for(int i=1; i<=N; ++i)    printf("%d
",Ans[i]%100003);//注意求模 
    return 0;
}