Luogu2164 SHOI2007 交通网络 期望、BFS、拓扑排序

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题目还算不难吧

首先我们枚举点$i$，将其他所有点到这个点的最短路求出来

然后我们在这一次建出的最短路$DAG$的反图上进行拓扑排序。假设我们算到了点$j$，点$j$的人流量为$t_j$，点$j$连出去的边到达的点为点集$\{v\}$，那么对于每一个点$u \in \{v\}$，边$(j,u)$的流量就会增加$\frac{t_j}{|\{v\}|}$，$t_u$会加上$\frac{t_j}{|\{v\}|}$

总时间复杂度$O(N^2)$

 #include<bits/stdc++.h>
 #define ld long double
 using namespace std;
 inline int read(){
     ;
     char c = getchar();
     while(!isdigit(c))    c = getchar();
     ) + (a << ) + (c ^ ') , c = getchar();
     return a;
 }
 inline int max(int a , int b){
     return a > b ? a : b;
 }
 struct Ed{
     int start , end , upEd;
 }ans[];
 ] , ifRail[][];
 ];
 ] , minRoute[];
 ld peo[][] , To[][];
 struct cmp{
     bool operator() (const int& a, const int& b ){
         return minRoute[a] < minRoute[b];
     }
 };
 int main(){
     int N = read() , M = read();
      ; i <= M ; i++){
         int a = read() , b = read();
         ans[(i << ) - ].start = a;
         ans[(i << ) - ].end = b;
         ans[(i << ) - ].upEd = firEd[a];
         firEd[a] = (i << ) - ;
         ans[i << ].start = b;
         ans[i << ].end = a;
         ans[i << ].upEd = firEd[b];
         firEd[b] = i << ;
         ifRail[a][b] = ifRail[b][a] = ;
     }
      ; i <= N ; i++)
          ; j <= N ; j++)    To[i][j] = read();
      ; i <= N ; i++){
         memset(minRoute , 0x3f , sizeof(minRoute));
         memset(Times ,  , sizeof(Times));
         minRoute[i] = ;
         Times[i] = ;
         queue < int > q;
         priority_queue < int , vector < int > , cmp > q1;
         q.push(i);
         while(!q.empty()){
             int t = q.front();
             q.pop();
             ;
             for(int j = firEd[t] ; j ; j = ans[j].upEd)
                 ){
                     minRoute[ans[j].end] = minRoute[t] + ;
                     Times[ans[j].end] = Times[t];
                     q.push(ans[j].end);
                     f = ;
                 }
                 ){
                     Times[ans[j].end] += Times[t];
                     f = ;
                 }
             if(!f)    q1.push(t);
         }
         memset(vis ,  , sizeof(vis));
         vis[i] = ;
         while(!q1.empty()){
             int t = q1.top();
             q1.pop();
              ; j <= N ; j++)
                 ){
                     ld t1 = (ld)To[i][t] * Times[j] / Times[t];
                     peo[j][t] += t1;
                     peo[t][j] += t1;
                     To[i][j] += t1;
                     if(!vis[j]){
                         vis[j] = ;
                         q1.push(j);
                     }
                 }
         }
     }
      ; i <= M ; i++)
         cout << ) << peo[ans[i << ].start][ans[i << ].end] + 1e- << endl;
     ;
 }