Poj 2135 farm tour (minimum fee Stream)

// Minimum fee stream <br/> // The Key to the diagram is to add a super source and a super collection point <br/> // you need to find a way to make a round trip, the sum of the two paths is the shortest and each path can only go through one <br/> // The first step is to find two paths from the start point to the end point <br/> // The second links to the billing flow, the distance is used as the cost. The capacity is set to 1 to ensure that each route passes only once <br/> // The capacity of the Super source point and the start point is 2, the capacity of the destination vertex and super vertex is 2, so that a maximum of 2 routes can be found <br/> // The final mincost is the answer <br/> // pay attention to the bidirectional edge condition, when two or more sides exist, <br/> # include <iostream> <br/> # include <br/> <queue> <br/> using namespace STD; <br/> const int maxn = 1003; <br/> const int maxm = 40005; <br/> const int INF = 1 000000000; <br/> int mincost, maxflow; <br/> int n, m; <br/> int U [maxm], V [maxm], Cap [maxm], flow [maxm], cost [maxm], next [maxm]; <br/> int head [maxn], pre [maxn], edge [maxn], DIS [maxn]; <br/> int m; <br/> void addedge (int u, int V, int cap, int cost) <br/> {<br/> flow [m] = flow [M + 1] = 0; <br/> V [m] = V; <br/> U [m] = u; <br/> CAP [m] = CAP; <br/> cost [m] = cost; <br/> next [m] = head [u]; <br/> head [u] = m ++; </P> <p> V [m] = u; <B R/> U [m] = V; <br/> CAP [m] = 0; <br/> cost [m] =-cost; <br/> next [m] = head [v]; <br/> head [v] = m ++; <br/>}< br/> void buildgraph () <br/>{< br/> scanf ("% d", & N, & M); <br/> int U, V, C; <br/> m = 0; <br/> memset (Head,-1, sizeof (head); <br/> while (M --) <br/>{< br/> scanf ("% d", & U, & V, & C); <br/> addedge (u, v, 1, C); <br/> addedge (v, U, 1, C); <br/>}< br/> addedge (0, 1, 2, 0 ); <br/> addedge (n, n + 1, 2, 0); <br/>}< br/> void mcmf (Int St, int ed) <br/>{< br/> queue <int> q; <br/> memset (flow, 0, sizeof (flow )); <br/> mincost = maxflow = 0; <br/> for (;) <br/> {<br/> bool INQ [maxn]; <br/> for (INT I = sT; I <= Ed; ++ I) dis [I] = (I = sT? 0: INF); <br/> memset (INQ, 0, sizeof (INQ); <br/> q. Push (ST); <br/> while (! Q. empty () <br/>{< br/> int u = Q. front (); q. pop (); <br/> INQ [u] = 0; <br/> for (int e = head [u]; e! =-1; E = next [e]) <br/>{< br/> If (Cap [e]> flow [e] & dis [u] + cost [e] <dis [V [e]) <br/> {<br/> dis [V [e] = dis [u] + cost [E]; <br/> pre [V [e] = U [E]; <br/> edge [V [e] = E; <br/> If (! INQ [V [e]) <br/>{< br/> q. push (V [e]); <br/> INQ [V [e] = 1; <br/>}< br/>}// spfa extended <br/> If (DIS [ed] = inf) break; <br/> int Delta = inf; // Delta can be improved <br/> for (INT u = Ed; u! = ST; u = pre [u]) <br/> Delta = min (delta, Cap [edge [u]-flow [edge [u]); // traverse the edges of the shortest path and modify the number of improvements <br/> for (INT u = Ed; u! = ST; u = pre [u]) <br/>{< br/> flow [edge [u] + = delta; // update forward traffic <br/> flow [edge [u] ^ 1]-= delta; // obtain the reverse edge number through an exception or 1, and updates reverse traffic <br/>}< br/> mincost + = dis [ed] * delta; <br/> maxflow + = delta; <br/>}</P> <p> int main () <br/>{< br/> // freopen ("in.txt ", "r", stdin); <br/> buildgraph (); <br/> mcmf (0, n + 1 ); <br/> printf ("% d/N", mincost); <br/> return 0; <br/>}