Dijkstra的Algo實現使用STL

Question

#include<bits/stdc++.h> 
using namespace std; 
#define ll long long 
vector<pair<ll,ll> >v[100005]; 
ll dis[100005]; 
bool visited[100005]; 
multiset<pair <ll,ll> > s; 
int main(){ 
    ll n,m,from,next,weight,i; 
    cin>>n>>m; 
    for(i=1;i<=n;i++){ 
     v[i].clear(); 
     dis[i]=2e9; 
    } 
    for(i=1;i<=m;i++){ 
     cin>>from>>next>>weight; 
     v[from].push_back(make_pair(next,weight)); 
     v[next].push_back(make_pair(from,weight)); 
    } 
    dis[1]=0; 
    s.insert({0,1}); 
    memset(visited,false,sizeof(visited)); 
    while(!s.empty()){ 
     pair<ll,ll>p= *s.begin(); 
     s.erase(s.begin()); 
     ll x=p.second; 
     ll wei=p.first; 
     if(visited[x]) continue; 
     for(i=0;i<v[x].size();i++){ 
      ll e=v[x][i].first; 
      ll w=v[x][i].second; 
      if(dis[x]+w < dis[e]){ 
       dis[e]=dis[x]+w; 
       s.insert({dis[e],e}); 
      } 
     } 
    } 
    for(i=2;i<=m;i++) 
    cout<<dis[i]<<" "; 
} 

我有一個c + +實現Dijkstra的Algo，但我想這不適用於所有情況下（較大的測試用例）。誰能幫我解決這個問題嗎？我是否遺漏了某些東西或者沒有實施。 該代碼輸出來自源頂點（即1）的每個頂點的最小距離。

Answer 1

您從不寫信給visited陣列。因此邊緣可能會被多次掃描。簡單的解決：在if(visited[x]) continue;後添加一行：

visited[x] = true; 

Answer 2

這是我的解決方案爲O解決（N）的曲線圖： 的#include 的#include 的#include

 typedef long long ll; 
    void fs_int(int *x) { 
     register int c = getchar_unlocked(); 
     *x = 0; 
     int neg = 0; 

     for(; ((c<48 || c>57) && c != '-'); c = getchar_unlocked()); 

     if(c=='-') { 
      neg = 1; 
      c = getchar_unlocked(); 
     } 

     for(; c>47 && c<58 ; c = getchar_unlocked()) { 
      *x = (*x<<1) + (*x<<3) + c - 48; 
     } 

     if(neg) 
      *x = -(*x); 
    } 
    typedef struct { 
     int next; 
     int val; 
     int d; 

    }List; 
    typedef struct 
    { 
     int parent; 
     int shrt; 
     ll count; 
     int on_reg; 
     int ch; 
    } Node; 
    #define MOD 1000000007 
    ll get_sum(Node *tr,List *l) 
    { 
     Node *t, *t2; 
     int i,j,n=0,fix; 
     ll result; 
     static int *reg=NULL,sz=1000; 

     if (!reg) 
      reg=malloc(sizeof(int)*sz); 
     reg[n++]=1; 
     int cur_d; 

     while(n) 
     { 
      ///fix is the limit for the for, it is the shortname of "for ix" : 
      // from 0 to fix there are the old values, from fix to n there are the new ones 
      fix=n; 
      for (i=0;i<fix;i++) 
      { 

       //the better way to reduce the complexity is shift the last item to the current one 
       t=&tr[reg[i]]; 
       reg[i--]=reg[--fix]; 
       reg[fix]=reg[--n]; 
       t->on_reg=0; 

       ///this scores all the edges from departing from this node 
       ///the criteria to avoid propagation is the key of the program 
       for (j=t->ch;j;j=l[j].next) 
       { 
        if (l[j].val==1) //avoid the root 
         continue; 

        t2=&tr[l[j].val]; //store in some comfortable variable 

        cur_d=t->shrt+l[j].d; 

        if (t2->shrt!=0 && t2->shrt< cur_d) ///if my path is heaviest nothing to do 
         continue; 
        else if (t2->shrt ==cur_d) //I found an item with same weight. It was required to count them 
         t2->count++; 
        else if (t2->shrt==0 || t2->shrt>cur_d) //found a unexplored item or my path is lighter 
        { 
         t2->shrt=cur_d; 
         t2->count=1; 
         if (!t2->on_reg) //if not already in the reg, I insert it inside 
         { 
          if (n>=sz) 
          { 
           sz<<=1; 
           reg=realloc(reg, sizeof(int)*sz); 
          } 
          reg[n++]=l[j].val; //at position n 
          t2->on_reg=1; 
         } 
        } 

       } 
      /* printf ("reg: "); 
      for (k=0;k<n;k++) 
       printf ("%d ",reg[k]); 
       printf ("\n");*/ 
      } 
     } 

     //printf ("\n"); 
     return result; 


    } 

    typedef long long ll; 
    void set_depth(Node *tr, List *l, int rt,int cd,int parent) 
    { 
     int i; 

     tr[rt].parent=parent; 
     for (i=tr[rt].ch;i;i=l[i].next) 
      if (l[i].val== parent) 
       continue; 
      else 
       set_depth(tr,l,l[i].val,cd+1,rt); 
    } 

    int main() 
    { 

     int t,n,q,i,u,v,d; 
     fs_int(&t); 
     int il=1; 
     Node tr[100005]; 
     List l[200005]; 
     List *tl; 
     while (t--) 
     { 
      fs_int(&n); 
      fs_int(&q); 
      il=1; 

      memset(tr,0,sizeof(tr)); 
      memset(l,0,sizeof(l)); 

      for (i=0;i<q;i++) 
      { 
       fs_int(&u); 
       fs_int(&v); 
       fs_int(&d); 

       tl=&l[il]; 
       tl->next=tr[u].ch; 
       tl->val=v; 
       tl->d=d; 
       tr[u].ch=il++; 


       tl=&l[il]; 
       tl->next=tr[v].ch; 
       tl->val=u; 
       tl->d=d; 
       tr[v].ch=il++; 

      } 

      //set_depth(tr,l,1,0,0); 
      // print(tr,l,1,0,0); 

      get_sum(tr,l); 

      ll res=1; 
      for (i=2;i<=n;i++) 
      { 


       res= ((res%MOD) *(tr[i].count%MOD))%MOD; 
      } 
      printf ("%lld\n",res); 
     } 

     return 0; 
    } 

功能你感興趣的是函數get_sum（）。這是一個廣度優先搜索，在圖中意味着檢查同心圓，這可以避免無用的傳播。它將這個虛擬圓中的值存儲在一個名爲reg的數組中。在每一步你都要檢查。關於效率，你可以在Ways比賽中自我檢查。它有一個最好的時間