題目連結：https://www.acwing.com/problem/content/385/

次短距離一定只能由次短距離更新

程式碼：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include<vector>
using namespace std;
const int N = 2010,M=2e5+10;
int t,m,n;
int h[N], e[M], ne[M], w[M],idx;
int dist[N][2],st[N][2],cnt[N][2];//拆點，將一個點拆成最短長度和次短長度兩個，當成兩個點分別加入dij的pq中
int S,T;
void add(int a, int b,int c) // 新增一條邊a->b
{
  e[idx] = b, ne[idx] = h[a],w[idx]=c, h[a] = idx ++ ;
}
struct ver{
  int id,type,dist;
  bool operator>(const ver&other)const{
    return dist>other.dist;
   }
};
int dij(){
  dist[S][0]=0;
  cnt[S][0]=1;
  priority_queue<ver,vector<ver>,greater<ver> > pq;
  pq.push({S,0,0});
  while(pq.size()){
    auto cur=pq.top();
    pq.pop();
    int id=cur.id,type=cur.type,distance=cur.dist,count=cnt[id][type];
    if(st[id][type])continue;
    st[id][type]=true;
    for(int i=h[id];~i;i=ne[i]){
      int k=e[i];
      if(dist[k][0]>distance+w[i]){//如果更新了最短路，那次短路也要更新，注意注意
        dist[k][1]=dist[k][0];
        dist[k][0]=distance+w[i];
        cnt[k][1]=cnt[k][0];
        cnt[k][0]=count;
        pq.push({k,0,dist[k][0]});
        pq.push({k,1,dist[k][1]});//要同時加入兩個點，因為兩個點都被更新過
       }
      else if(dist[k][0]==distance+w[i])cnt[k][0]+=count;
      else if(dist[k][1]>distance+w[i]){
        dist[k][1]=distance+w[i];
        cnt[k][1]=count;
        pq.push({k,1,dist[k][1]});
       }
      else if(dist[k][1]==distance+w[i])cnt[k][1]+=count;
     }
   }
  int res=cnt[T][0];
  if(dist[T][1]==dist[T][0]+1)res+=cnt[T][1];
  return res;
}
​
int main(){
  cin>>t;
  while(t--){
    memset(h, -1, sizeof h);
    idx=0;
    memset(dist,0x3f,sizeof dist);
    memset(st, 0, sizeof st);
    memset(cnt,0,sizeof cnt);
    cin>>n>>m;
    while(m--){
      int a,b,c;
      cin>>a>>b>>c;
      add(a,b,c);
     }
    cin>>S>>T;
    cout<<dij()<<endl;
  
   }
}