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Topic:

John has n plots, home at number 1th, and N is a warehouse. There are M-roads in the farm (bidirectional), road I connects the AI and bi, and the length is CI.

John wanted to go to the warehouse after a few visits from home. And then go back home. Suppose that a round trip cannot go through the same road two times, the minimum value of the total length of the route is obtained.

Algorithm Analysis:

Solve with shortest path and then delete the edge in the first minimum to solve a short circuit at a time. Is this feasible? A counter-example should be found immediately to prove that the method does not always get the best results.

So we give up the idea of the question as a go and back, and instead use the question as a way to find two without a common edge from vertex 1th to n vertex? After this conversion. is not to ask for a flow of 2 of the minimum cost, because the road is bidirectional.

#include <iostream> #include <algorithm> #include <vector> #include <queue> #include <cstdio > #include <cstring>using namespace std;/* flow limit is F. Solving the minimum cost time complexity: O (F mlogn) */typedef pair<int,int> p;const int INF = 1 << 30;const int maxn = + + 10;str   UCT edge{int To,cap,cost,rev;   Edge () {}; Edge (int _to,int _cap,int _cost,int _rev): to (_to), Cap (_CAP), Cost (_cost), rev (_rev) {};};        int V;     Vertex int n,m,s,t;vector<edge> G[maxn];int H[MAXN];  The potential of the vertex int DIST[MAXN];  Shortest distance int PREVV[MAXN],PREVE[MAXN]; The shortest path in the German precursor node and the corresponding side void init () {S = 1; T = N;   V = T + 1; for (int i = 0;i <= v;++i) g[i].clear ();} Add an edge void Addedge (int from,int to,int cap,int cost) {G[from].push_back (edge) (to,cap,cost,g[to    ].size ())); G[to].push_back (Edge (From,0,-cost,g[from].size ()-1));} Solve the minimum cost flow from S to T flow to f//assume no flow for F.    Then return -1int min_cost_flow (int s,int t,int f) {//s: start T: End F: Flow limit int res = 0; Fill (h,h +v,0); Initialize h while (F > 0) {//update H PRIORITY_QUEUE&LT;P,VECTOR&LT;P&GT;,GREATER&LT;P&GT using Dijkstra algorithm ;        > Q;        Fill (dist,dist + v,inf);        Dist[s] = 0;        Q.push (P (0,s)); while (! Q.empty ()) {p p = q.top ();            Q.pop ();            int v = p.second;            if (Dist[v] < P.first) continue;                for (int i = 0;i < G[v].size (); ++i) {edge& e = g[v][i];                int tmp = Dist[v] + e.cost + h[v]-h[e.to];                    if (E.cap > 0 && dist[e.to] > tmp) {dist[e.to] = tmp;                    Prevv[e.to] = v;                    Preve[e.to] = i;                Q.push (P (dist[e.to],e.to));        }}}//while//cannot be augmented if (dist[t] = = INF) {return-1;        } for (int v = 1;v <= v;++v) h[v] + = Dist[v];        int d = f; for (int v = t;v! = S;v = Prevv[v]) {d = min (d,g[prevv[V]]        [Preve[v]].cap];        } F-= D;        Res + = d * H[t];            for (int v = t; v! = S;v = Prevv[v]) {edge& e = g[prevv[v]][preve[v]];            E.cap-= D;        G[v][e.rev].cap + = D; }} return res;}    int main () {//freopen ("Input.txt", "R", stdin);        while (~SCANF ("%d%d", &n,&m)) {init ();        int a,b,c;            for (int i = 0;i < M;++i) {scanf ("%d%d%d", &a,&b,&c);            Addedge (A,B,1,C);        Addedge (B,A,1,C);    } printf ("%d\n", Min_cost_flow (s,t,2)); } return 0;}

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