Graph theory $\cdot$ Shortest circuit problem

• Dijkstra single source Shortest path algorithm

The Dijkstra can calculate the shortest path from the starting point to each point, and the single source minimum (SSSP). This feature makes Dijkstra often used for preprocessing of other algorithms. The code to calculate the shortest path with the Dijkstra algorithm is as follows:

Note: Code comments See the Getting started with algorithmic competition-training Guide (Rujia)

1 structdijkstra{2     intN, M;3Vector<e>e;4vector<int>G[MAXN];5     BOOLDONE[MAXN];6     intD[MAXN];7     intP[MAXN];8     voidInitintN) {9          This->n =N;Tenfor (I,0N1) g[i].clear (); One e.clear (); A     } -     voidAdde (int  from,intTo,intDist) { -E.PB (E ( from, to, Dist)); them =e.size (); -g[ from].PB (M-1); -     } -     voidDijkstraints) { +Priority_queueQ; -for (I,0N1) D[i] =Int_inf; +D[s] =0; ACLR (Done,0); atQ.push (Heapnode (0, s)); -          while(!Q.empty ()) { -Heapnode x =Q.top (); Q.pop (); -             intU =x.u; -             if(Done[u])Continue; -Done[u] =1; in             intSZ =g[u].size (); -for (I,0, Sz-1){ toE &y =E[g[u][i]]; +                 if(D[y.to] > D[u] +y.dist) { -D[y.to] = D[u] +y.dist; theP[y.to] =G[u][i]; * Q.push (Heapnode (d[y.to], y.to)); $                 }Panax Notoginseng             } -         } the     } +};

• Bellman-ford algorithm

An important application of the Bellman-ford algorithm is to award negative loops. If the relaxation (relax) operation can be performed after the iteration $n-1$, there must be a negative circle. If the queue is implemented, then when a node is enqueued $n$ times can be determined that there is a negative circle, the code is as follows:

1 structbellman_ford{2     intN, M;3Vector<e>e;4vector<int>G[MAXN];5     BOOLINQ[MAXN];6     intD[MAXN];7     intP[MAXN];8     intCNT[MAXN];9     voidInitintN) {Ten          This->n =N; Onefor (I,0N1) g[i].clear (); A e.clear (); -     } -     voidAdde (int  from,intTo,intDist) { theE.PB (E ( from, to, Dist)); -m =e.size (); -g[ from].PB (M-1); -     } +     BOOLNegcyc () { -queue<int>Q; +CLR (INQ,0), CLR (CNT,0); Afor (I,0N1) D[i] =0, inq[i] =1, Q.push (i); at          while(!Q.empty ()) { -             intU =Q.front (); Q.pop (); -Inq[u] =0; -             intSZ =g[u].size (); -for (I,0, Sz-1){ -E &y =E[g[u][i]]; in                 if(D[y.to] > D[u] +y.dist) { -D[y].to = D[u] +y.dist; toP[e.to] =G[u][i]; +                     if(!Inq[y.to]) { - Q.push (y.to); theInq[y.to] =1; *                         if(++cnt[y.to] > N)return 1; $                     }Panax Notoginseng                 } -             } the         } +         return 0; A     } the};

Graph theory $\cdot$ Shortest circuit problem