題意

求最短路徑的數目，可能用重邊。

題解

SPFA，為了保證不會重複計算，需要維護一個當前佇列中到這些頂點的次數，每次出隊時清零。

程式碼

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 4e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f
 
;
const ull p = 67;
const ull MOD = 1610612741;
int n, m, tot;
int dist[nmax], cnt[nmax], head[nmax], thiscnt[nmax];
int mp[2005][2005];
int inque[nmax];
struct edge{
    int to, nxt, w;
}e[nmax<<1];
void add_edge(int u, int v, int w) {
    e[tot].to = v;
    e[tot].nxt = head[u];
    e[tot].w = w;
    head[u] = tot ++;
}
void
 
 spfa(int s) {
    for(int i = 1; i <= n; ++i) {
        dist[i] = i == s? 0 : INF;
        inque[i] = false;
    }
    queue<int> q;
    q.push(s);
    inque[s] = true;
    thiscnt[s] = cnt[s] = 1;
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        inque[u] = false;
        if
 
(u == n)
            continue;
        for(int v = 1; v <= n; ++v) {
            if(u == v || mp[u][v] == 0)
                continue;
            if(dist[v] > dist[u] + mp[u][v]) {
                dist[v] = dist[u] + mp[u][v];
                cnt[v] = thiscnt[v] = thiscnt[u];
                if(!inque[v]) {
                    inque[v] = true;
                    q.push(v);
                }
            } else if(dist[v] == dist[u] + mp[u][v]) {
                cnt[v] += thiscnt[u];
                thiscnt[v] += thiscnt[u];
                if(!inque[v]) {
                    inque[v] = true;
                    q.push(v);
                }
            }
        }
        thiscnt[u] = 0;

    }
}
//void dijkstra(int s) {
//
//}
int main() {
    scanf("%d %d",&n, &m);
    memset(head, -1, sizeof head);
    int u, v, w;
    for(int i = 1;i <= m; ++i) {
        scanf("%d %d %d", &u, &v, &w);
        if(mp[u][v] == 0)
            mp[u][v] = w;
        else
            mp[u][v] = min(mp[u][v], w);
//        add_edge(u,v,w);
//        add_edge(v,u,w);
    }
    spfa(1);
    if(dist[n] == INF) {
        printf("No answer\n");
    } else {
        printf("%d %d\n", dist[n], cnt[n]);
    }
    return 0;
}