P2505 [HAOI2012] 道路

题意

\(n\) 个点，\(m\) 条边的带权有向图，问每条边被包含在几条最短路中。
\(n\le1500,m\le5000,1\le w\le10^4\)/

思路

枚举一个起点，跑最短路，求出到每个点的最短路数量，从起点到点 \(i\) 的最短路数量记为 \(g_i\)，长度记为 \(dis_i\)。建一张新图 \(G\)，这张图上的边 \((u,v,w)\) 满足 \(dis_u+w=dis_v\)，且满足条件的边全部在这张图里。显然，这是一只 DAG。按照拓扑序遍历整个图，设 \(f_i\) 表示从 \(i\) 号点开始，在 \(G\) 中走，任意点停下的方案数。\(f_i=1+\sum_{(i,v,w)\in G} f_v\)，对于一条边 \((u,v,w)\in G\)，祂在当前起点的方案数为 \(g_u\times f_v\)。

代码

/*
Luogu P2505 [HAOI2012] 道路
2026-04-15
*/
#include<bits/stdc++.h>
using namespace std;
namespace IO{
    template<typename T>
    inline void read(T&x){
        x=0;char c=getchar();bool f=0;
        while(!isdigit(c)) c=='-'?f=1:0,c=getchar();
        while(isdigit(c)) x=x*10+c-'0',c=getchar();
        f?x=-x:0;
    }
    template<typename T>
    inline void write(T x){
        if(x==0){putchar('0');return ;}
        x<0?x=-x,putchar('-'):0;short st[50],top=0;
        while(x) st[++top]=x%10,x/=10;
        while(top) putchar(st[top--]+'0');
    }
    inline void read(char&c){c=getchar();while(isspace(c)) c=getchar();}
    inline void write(char c){putchar(c);}
    inline void read(string&s){s.clear();char c;read(c);while(!isspace(c)&&~c) s+=c,c=getchar();}
    inline void write(string s){for(int i=0,len=s.size();i<len;i++) putchar(s[i]);}
    template<typename T>inline void write(T*x){while(*x) putchar(*(x++));}
    template<typename T,typename...T2> inline void read(T&x,T2&...y){read(x),read(y...);}
    template<typename T,typename...T2> inline void write(const T x,const T2...y){write(x),putchar(' '),write(y...),sizeof...(y)==1?putchar('\n'):0;}
}using namespace IO;
template<int mod>struct Modint{
    int z;
    Modint(){z=0;}
    Modint(int x){x%=mod;z=x<0?x+mod:x;}
    Modint(long long x){x%=mod;z=x<0?x+mod:x;}
    Modint(short x){x%=mod;z=x<0?x+mod:x;}
    Modint(char x){x%=mod;z=x<0?x+mod:x;}
    Modint(bool x){x%=mod;z=x<0?x+mod:x;}
    friend Modint operator+(Modint t,Modint t2){Modint ans;ans.z=(t.z+t2.z)%mod;return ans;}
    friend Modint operator*(Modint t,Modint t2){Modint ans;ans.z=1ll*t.z*t2.z%mod;return ans;}
    friend Modint operator-(Modint t,Modint t2){Modint ans;ans.z=(t.z-t2.z)%mod;return ans;}
    Modint operator<<(const int t)const{Modint ans;ans.z=(z<<t)%mod;return ans;}
    Modint operator>>(const int t)const{Modint ans;ans.z=(z>>t)%mod;return ans;}
    Modint&operator+=(const Modint t){z=(z+t.z)%mod;return *this;}
    Modint&operator*=(const Modint t){z=1ll*z*t.z%mod;return *this;}
    Modint&operator-=(const Modint t){z=(z-t.z)%mod;return *this;}
    Modint&operator<<=(const int t){z=(z<<t)%mod;return *this;}
    Modint&operator>>=(const int t){z=(z>>t)%mod;return *this;}
    Modint&operator++(){z++,z%=mod;return *this;}
    Modint&operator--(){z--,z%=mod;return *this;}
    Modint operator++(int){Modint ls=*this;z++,z%=mod;return ls;}
    Modint operator--(int){Modint ls=*this;z--,z%=mod;return ls;}
    friend Modint ksm(Modint a,int b){
        Modint ans=1;
        while(b){if(b&1) ans=ans*a;a=a*a,b>>=1;}
        return ans;
    }
    friend void read(Modint&z){
        int x=0;char c=getchar();bool f=0;
        while(!isdigit(c)) c=='-'?f=1:0,c=getchar();
        while(isdigit(c)) x=(x*10ll+c-'0')%mod,c=getchar();
        f?x=-x:0;
        z.z=x;
    }
    friend void write(Modint x){x.z<0?x.z+=mod:0;write(x.z);}
};
const int maxn=1510,maxm=5010,inf=1000000000,mod=1000000007;
#define M Modint<mod>
int n,m,dis[maxn],d[maxn];
M f[maxn],g[maxn],ans[maxm];
bool flag[maxn];
struct EDGE{int u,v,w;}b[maxm];
vector<pair<int,int>>e[maxn];
vector<int>e2[maxn];
priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>q;
void dij(int be){
    for(int i=1;i<=n;i++) dis[i]=inf,flag[i]=0,g[i]=0;
    q.push({0,be}),dis[be]=0,g[be]=1;
    while(!q.empty()){
        auto[d,u]=q.top();q.pop();
        if(flag[u]) continue;flag[u]=1;
        for(auto[v,w]:e[u]){
            if(dis[v]<dis[u]+w) continue;
            if(dis[v]==dis[u]+w){g[v]+=g[u];continue;}
            dis[v]=dis[u]+w,g[v]=g[u];
            q.push({dis[v],v});
        }
    }
}
signed main(){
    read(n,m);
    for(int i=1;i<=m;i++){
        int u,v,w;read(u,v,w);
        b[i]={u,v,w};
        e[u].push_back({v,w});
    }
    for(int i=1;i<=n;i++){
        dij(i);
        for(int i=1;i<=n;i++) e2[i].clear(),f[i]=1;
        for(int u=1;u<=n;u++) for(auto[v,w]:e[u]) if(dis[v]==dis[u]+w) e2[v].push_back(u),d[u]++;
        queue<int>q;
        for(int j=1;j<=n;j++) if(d[j]==0) q.push(j);
        while(!q.empty()){
            int u=q.front();q.pop();
            for(int v:e2[u]){
                if(--d[v]==0) q.push(v);
                f[v]+=f[u];
            }
        }
        for(int j=1;j<=m;j++){
            auto[u,v,w]=b[j];
            if(dis[u]+w==dis[v]) ans[j]+=g[u]*f[v];
        }
    }
    for(int i=1;i<=m;i++) write(ans[i]),write("\n");
    return 0;
}