[洛谷1144]最短路计数

思路：

Dijkstra求最短路，同时统计最短路个数，当新增最短路时，增加的条数是父亲结点的条数。

#include<cstdio>
#include<cctype>
#include<vector>
#include<functional>
#include<ext/pb_ds/priority_queue.hpp>
inline int getint() {
    char ch;
    while(!isdigit(ch=getchar()));
    int x=ch^'0';
    while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
    return x;
}
const int N=1000000,inf=0x7fffffff,mod=100003;
std::vector<int> e[N];
inline void add_edge(const int u,const int v) {
    e[u].push_back(v);
    e[v].push_back(u);
}
struct Vertex {
    int id,d,ans;
    bool operator > (const Vertex &x) const {
        return d>x.d;
    }
};
__gnu_pbds::priority_queue<Vertex,std::greater<Vertex> > q;
__gnu_pbds::priority_queue<Vertex,std::greater<Vertex> >::point_iterator p[N];
bool v[N]={0};
int main() {
    int n=getint(),m=getint();
    while(m--) {
        add_edge(getint(),getint());
    }
    for(int i=1;i<=n;i++) {
        p[i]=q.push((Vertex){i,(i==1)?0:inf,(i==1)?1:0});
    }
    for(Vertex u=q.top();u.d!=inf;u=q.top()) {
        for(unsigned int i=0;i<e[u.id].size();i++) {
            if(v[e[u.id][i]]) continue;
            int &v=e[u.id][i];
            if(u.d+1<p[v]->d) {
                q.modify(p[v],(Vertex){v,u.d+1,p[u.id]->ans%mod});
            }
            else if(u.d+1==p[v]->d) {
                q.modify(p[v],(Vertex){v,p[v]->d,(p[v]->ans+p[u.id]->ans)%mod});
            }
        }
        q.modify(p[u.id],(Vertex){u.id,inf,u.ans});
        v[u.id]=true;
    }
    for(int i=1;i<=n;i++) {
        printf("%d
",p[i]->ans);
    }
    return 0;
}