1658-admiral (split + minimum charge flow)

The split-point method in this problem is a general method to solve several capacity. Because only the capacity limit is still unable to satisfy each node access only once this limit, the reason is very simple, we draw a diagram to know, assume that there are two paths from the starting point to the same node 2, and then all to the end of N, although they meet the traffic limit but through the same node.

So how do we solve this problem? The answer is: the Split method.

A node is split into two nodes, with a true junction with a capacity of 1 to 0 of the edge to the false junction, so that when we add edge, the other starting node is a false node, the end point is the true node. This implicitly adds a capacity attribute to this node. Of course, since we have to go through the starting point and the last point two times, so we can only 2~n-1 node, 1 and N to special treatment. If 1 or n is the starting point of the edge, then do not use false junction.

See the code for details:

#include <bits/stdc++.h>using namespace Std;typedef long long ll;const int INF = 100000000;const int maxn = 3*1000+5;    int n,m,u,v,c,t,p[maxn],a[maxn],inq[maxn],d[maxn];struct Edge {int from, to, cap, flow, cost; Edge (int u,int v,int c,int f,int W): From (U), to (v), Cap (c), Flow (f), Cost (W) {}};vector<edge> Edges;vector<int    > G[maxn];void init () {for (int i=0;i<maxn;i++) g[i].clear (); Edges.clear ();}    void Addedge (int from, int to, int cap, int. cost) {Edges.push_back (Edge (from,to,cap,0,cost));    Edges.push_back (Edge (to,from,0,0,-cost));    t = edges.size ();    G[from].push_back (t-2); G[to].push_back (t-1);}    BOOL Bellmanford (int s,int t,int& flow, ll& cost) {for (int i=0;i<maxn;i++) d[i] = INF;    memset (inq,0,sizeof (INQ)); D[s] = 0; Inq[s] = 1; P[s] = 0;    A[s] = INF;    Queue<int> Q;    Q.push (s); while (! Q.empty ()) {int u = q.front ();        Q.pop ();        Inq[u] = 0; for (int i = 0; i < g[u].size (); i++) {Edge& e = edges[g[u][i]];                if (E.cap > E.flow && d[e.to] > D[u] + e.cost) {d[e.to] = D[u] + e.cost;                P[e.to] = G[u][i];                A[e.to] = min (a[u],e.cap-e.flow);            if (!inq[e.to]) {Q.push (e.to); inq[e.to] = 1;}    }}} if (d[t] = = INF) return false;    Flow + = A[t];    Cost + = (LL) d[t] * (LL) a[t];        for (int u = t; u! = s; u = edges[p[u]].from) {Edges[p[u]].flow + = a[t];    Edges[p[u]^1].flow-= a[t]; } return true;    int mincostmaxflow (int s,int t, ll& cost) {int flow = 0; cost = 0;    while (Bellmanford (s,t,flow,cost)); return flow;}        int main () {while (~scanf ("%d%d", &n,&m) &&n) {init ();   for (int i=2;i<=n-1;i++) {Addedge (i,i+n,1,0);            The Split Method} for (int i=1;i<=m;i++) {scanf ("%d%d%d", &u,&v,&c);//The True node is connected by false nodes.            if (u! = 1 && u! = N) addedge (u+n,v,1,c); ElseAddedge (U,V,1,C);        } ll ans;        Addedge (0,1,2,0);        Addedge (n,2*n+1,2,0);        Mincostmaxflow (0,2*n+1,ans); printf ("%lld\n", ans);//i64d PE} return 0 on UVA;}

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1658-admiral (split + minimum charge flow)